HomeSecond Term ExaminationSenior Secondary Further Maths 1st term

Senior Secondary Further Maths 1st term

SUBJECT: FURTHER MATHEMATICS CLASS: SSS2

SCHEME OF WORK

WEEK TOPICS

  1. Polynomials 1: (a) Definition of polynomials (b) Basic operations on polynomials. (c) Remainder and factor theorem. (d) Zeros of polynomials.
  2. Polynomials 2: (a) Roots of cubic equation: Sum of roots; Sum of products of two Roots; Products of roots. (b) Graphs of polynomial function.
  3. Permutations: (a) Permutation (arrangement) (b) Cyclic permutation (c) Arrangement of identical objects. (c) Arrangement in which repetitions are allowed.
  4. Combination: (a) Combination (selection). (b) Conditional arrangements and selection. (c) Probability problems involving arrangement and selection.
  5. Binomial Expansion 1: (a) Pascal triangle (ii) Binomial expansion of (a+b)n , where n can be positive integer, negative integer or fractional value.
  6. Binomial Expansion 2: (a) Finding nth term (b) Application of binomial expansion
  7. Mid-term break.
  8. Roots of quadratic equation 1: (a) Quadratic equation (completing the square and formula method). (b) Sum and product of roots of quadratic equation (c) Finding quadratic equation given sum and product of roots, x2 – (sum of roots) + (product) = 0 (d) Condition for quadratic equation to have: (i) Equal roots (b2 = 4ac) (ii) Real roots b2 > 4ac (iii) No roots b2 < 4ac
  9. Roots of quadratic equation 2: (a) Conditions for given line to: (i) intersect a curve, (ii) be tangent to curve, (iii) not intersect a curve (b) Solution of problems on roots of quadratic equation. (c) Maximum and minimum values.
  10. Revision.
  11. Examination.

WEEK ONE

TOPIC: POLYNOMIALS 1

SUB-TOPICS:

(a) Definition of polynomials.

(b) Basic operations on polynomials.

(c) Remainder and factor theorem.

(d) Zeros of polynomials.

SUB-TOPIC 1

Definition of Polynomials

A polynomial is a mathematical expression which is a sum of terms, each term consisting a variable or variables raised to a power and multiplied by a coefficient. A polynomial of one variable x (univariate) has the following as its general form:

anxn + an-1xn-1 + … + a2x2 + a1x + a0

where the highest power of the variable n is the degree of the polynomial; the numerical constants an, an-1, … a2, a1 are called the coefficients of the polynomial, while a0 is called the constant term.

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Examples of polynomials include:

  • 3x2 – 2x + 4
  • 2x3 + 3x2 + 5x + 3

A function whose values are given by a polynomial is called a polynomial function. Eg: f(x) = 2x3 + 3x2 + 5x + 3

An equation that is obtained when we set a polynomial equal to zero is called a polynomial equation. E.g.: 2x3 + 3x2 + 5x + 3 = 0

Equality of polynomials

Two polynomials,

P(x) = anxn + an-1xn-1 + … + a2x2 + a1x + a0 and

Q(x) = bnxn + bn-1xn-1 + … + b2x2 + b1x + b0

are said to be equal if:

an = bn

an-1 = bn-1

a2 = b2

a1 = b1

a0 = b0

The value that is obtained by substituting a for x in a polynomial P(x) is denoted by P(a).

SUB-TOPIC 2

Basic operations on polynomial

Addition and Subtraction of Polynomials

Given

P(x) = anxn + an-1xn-1 + … + a2x2 + a1x + a0 and

Q(x) = bnxn + bn-1xn-1 + … + b2x2 + b1x + b0

Then,

P(x) + Q(x) = (an + bn) xn + (an-1 + bn-1) xn-1 + … + (a2 + b2) x2 + (a1 + b1) x + a0 + b0

Similarly,

P(x) + Q(x) = (an – bn) xn + (an-1 – bn-1) xn-1 + … + (a2 – b2) x2 + (a1 – b1) x + a0 – b0

Examples:

  1. Given P(x) = 5x3 – 3x2 + 4x + 7; Q(x) = 6x2 + 5x – 4; R(x) = 8x3 + 5x – 2.

Find, (a) P(x) + Q(x); (b) R(x) – P(x); (c) P(x) + 2Q(x) – 3R(x).

  1. If F(x) = 3x3 + 4x2 – 5x + 9, find: (a) F(-1). (b) F(1). (c) F(0). (d) F(3).

Solution:

  1. (a) P(x) + Q(x) = (5x3 – 3x2 + 4x + 7) + (6x2 + 5x – 4)

= 5x3 + (– 3x2 + 6x2) + (4x + 5x) + (7 + (-4))

= 5x3 + 3x2 + 9x + 3

(b) R(x) – P(x) = (8x3 + 5x – 2) – (5x3 – 3x2 + 4x + 7)

= (8x3 – 5x3) + (0 – (-3x2)) + (5x – 4x) + ((-2) -7)

= 3x3 + 3x2 + x – 9

(c) P(x) + 2Q(x) – 3R(x) = (5x3 – 3x2 + 4x + 7) + 2(6x2 + 5x – 4) – 3(8x3 + 5x – 2)

= (5x3 – 3x2 + 4x + 7) + (12x2 + 10x – 8) – (24x3 + 15x – 6)

= 5x3 – 3x2 + 4x + 7 + 12x2 + 10x – 8 – 24x3 – 15x + 6

= 5x3 – 24x3 – 3x2 + 12x2 + 4x + 10x – 15x + 7 – 8 + 6

= – 19x3 + 9x2 – x + 5

  1. F(x) = 3x3 + 4x2 – 5x + 9
  2. F(-1) = 3(-1)3 + 4(-1)2 – 5(-1) + 9

= 3(-1) + 4(1) – 5(-1) + 9

= -3 + 4 + 5 + 9

= 15

  1. F(1) = 3(1)3 + 4(1)2 – 5(1) + 9

= 3(1) + 4(1) – 5 + 9

= 3 + 4 – 5 + 9

= 11

  1. F(0) = 3(0)3 + 4(0)2 – 5(0) + 9

= 9

  1. F(3) = 3(3)3 + 4(3)2 – 5(3) + 9

= 3(27) + 4(9) – 15 + 9

= 81 + 36 – 15 + 9

= 111[mediator_tech][mediator_tech][mediator_tech]

Class activity

  1. Given that P1(x) = 2x2 + 3x + 4; P2(x) = 4x2 – 6x + 8; P3(x) = 5x3 – 3x2 + 5x + 6. Find 4 P1(x) + 5 P2(x) – 2 P3(x).
  2. If f(x) = x4 – 3x3 + x2 + 3x – 2, show that f(1) = f(2).

Multiplication of Polynomials

The multiplication of two polynomials is obtained by using every term of one polynomial to multiply each term of the other polynomial and collecting together like terms.

When a polynomial of degree m is multiplied by another polynomial of degree n, another polynomial of degree m + n is obtained.

Examples

  1. Given P(x) = 7x3 – 4x2 + 3x +4 and Q(x) = 5x2 + 6x +1. Find PQ.

Solution:

Method 1:

PQ = (7x3 – 4x2 + 3x +4)(5x2 + 6x +1)

= 7x3(5x2 + 6x +1) – 4x2(5x2 + 6x +1) + 3x(5x2 + 6x +1) +4(5x2 + 6x +1)

= 35x5 + 42x4 + 7x3 – 20x4 – 24x3 – 4x2 + 15x3 + 18x2 + 3x + 20x2 + 24x + 4

Rearrange

= 35x5 + 42x4 – 20x4 + 7x3 – 24x3 + 15x3 – 4x2 + 18x2 + 20x2 + 24x + 3x + 4

= 35x5 + 22x4 – 2x3 + 34x2 + 27x + 4

Method 2: (Long Multiplication)

7x3 – 4x2 + 3x +4

X 5x2 + 6x + 1

7x3 – 4x2 + 3x +4

42x4 – 24x3 + 18x2 + 24x

35x5 – 20x4 + 15x3 + 20x2

35x5 + 22x4 – 2x3 +34x2 + 27x +4

  1. Given that P(x) = 5x3 + 3x2 – 2x + 4 and Q(x) = 2x2 + 1, find P(x).Q(x)

Solution

Method 1:

PQ = QP = (2x2 + 1) (5x3 + 3x2 – 2x + 4)

= 2x2(5x3 + 3x2 – 2x + 4) + 1 (5x3 + 3x2 – 2x + 4)

= 10x5 + 6x4 – 4x3 + 8x2 + 5x3 + 3x2 – 2x + 4

= 10x5 + 6x4 + x3 +11x2 – 2x + 4

Method 2:

5x3 + 3x2 – 2x + 4

2x2 + 1

5x3 + 3x2 – 2x + 4

10x5 + 6x4 – 4x3 + 8x2

10x5 + 6x4 + 2x3 +11x2 – 2x +4

Class activity

  1. If f(x) = 3x2 – 2x + 3, g(x) = x2 – 3x + 4 and h(x) = x2 + 5, find f(x).g(x).h(x).
  2. Given that P1(x) = 5x3 + 3x2 – 2x + 6; P2(x) = x3 + 4x2 – 3x + 1 and P3(x) = 2x3 – 3x2 + 3x + 2, find (i) (P1 + P2)P3 (ii) (P2 – P3)P1.

Division of polynomials

A polynomial f(x) can be divided by another polynomial g(x), provided the degree of g(x) is not greater than that of f(x).

So, we could divide say x3 + 3x2 – x + 1 (3rd degree), by x-2 (1st degree), but not x – 2 by x3 + 3x2 –x + 1.

If n ≥ m, the result of dividing a polynomial of degree n by a polynomial of degree m is another polynomial of degree n – m.

Suppose we want to divide x3 + 3x2 – x + 1 by x-2, the polynomial x3 + 3x2 – x + 1is called the dividend while the polynomial x-2 is called the divisor. The result of the division is called the quotient and what is left after the division is called the remainder.

Examples

  1. Find the quotient and remainder when x3 + 3x2 –x + 1 is divided by x – 2.
  2. Find the quotient and remainder when 4x3 – x2 + x – 5 is divided by x2 + x – 1.

Solution 1.

Step 1: Divide the first term of the dividend by the first term of the divisor to obtain the first term of the quotient.

X2

x3 + 3x2 – x + 1

x + 1

X – 2

Step 2: Multiply the divisor by the first term of the quotient gotten and write the result under the dividend.

X2

x3 + 3x2 – x + 1

x3 – 2x2

x

X – 2

Step 3: Subtract the product obtained in step 2 from the dividend.

X2

x3 + 3x2 – x + 1

x3 – 2x2

5x2 – x + 1

x

X – 2

Step 4: Repeat steps 1, 2, and 3 with x2 – 5x + 1 as the new dividend.

X2 + 5x

X – 2

x3 + 3x2 – x + 1

x3 – 2x2

5x2 – x + 1

5x2–10x

9x + 1

x

Step 5: Repeat steps 1, 2, and 3 with 9x + 1 as the new dividend.

X2 + 5x + 9

x3 + 3x2 – x + 1

-(x3 – 2x2)

5x2 – x + 1

-(5x2–10x)

9x + 1

-(9x -18)

19

x

X – 2

 

Note: x3 + 3x2 – x + 1 is the dividend.

x – 2 is the divisor.

X2 + 5x + 9 is the quotient.

19 is the remainder.

We can combine the quotient, divisor and remainder to get the polynomial as follows:

x3 + 3x2 –x + 1 = (x – 2) x (x2 + 5x + 9) + 19

Polynomial (P) = Divisor (D) x Quotient (Q) + Remainder(R)

Solution 2

4x – 5

x2 + x – 1 4x3 – x2 + x – 5

-(4x3 + 4x2 – 4x)

  • 5x2 + 5x – 5

-(-5x2 – 5x + 5)

10x – 10

Note that in each of the examples above, the degree of Q = degree of P – degree of D and that the degree of R is one less than the degree of D. The degree of a remainder is one less than that of the divisor.

Class activity

  1. Find the quotients and remainders when P(x) = 6x3 + 4x2 – x + 5 is divided by:
  2. 3x2 + 2x +1
  3. x3 – 3x + 2
  4. A polynomial is divided by x + 1. The quotient is 2x – 3 and the remainder is 3, find the polynomial.

SUB-TOPIC 3

Remainder and factor theorem

The long division method used in the previous sub-topic helps us to determine, not only the quotient but also the remainder.

Consider the example 1 of the previous sub-topic:

Find the remainder when x3 + 3x2 –x + 1 is divided by x – 2.

X2 + 5x + 9

x3 + 3x2 – x + 1

-(x3 – 2x2)

5x2 – x + 1

-(5x2–10x)

9x + 1

-(9x -18)

19

x

X – 2

Hence when P(x) is divided by x – 2, the remainder is 19

Now find P(2).

P(2) = (2)3 + 3(2)2 – (2) + 1 = 8 + 12 – 2 + 1 = 19.

From the above, you would have observed that when p(x) is divided by x – a, the remainder is f(a). This forms the basis of Remainder theorem.

The Remainder theorem states that if a polynomial p(x) is divided by x – a, the remainder is p(a).

More generally, if p(x) is divided by ax + b, then the remainder is p(

A special case of the remainder theorem is when p(x) leaves no remainder when it is divided by x – a. when this happens, we say x – a is a factor of p(x).

This modified theorem is called factor theorem and it is states that if p(a) = 0 then x – a is a factor.

Examples

  1. Find the remainder when 2x2 – 5x + 6 is divided by x – 3.
  2. Determine the values of p and q if (x – 1) and (x + 2) are factor of 2x3 + px2 – x + q.

Solution:

  1. Let f(x) = 2x2 – 5x + 6

Let R be the remainder when f(x) is divided by x – 3

Then,

R = f(3)

F(3) = 2(3)2 – 5(3) + 6

= 2(9) – 15 + 6

= 18 – 15 + 6

= 9

  1. Let f(x) = 2x3 + px2 – x + q

If x – 1 is a factor of f(x), then

F(1) = 0

F(1) = 2(1)3 + p(1)2 – (1) + q

= 2 + p – 1 + q

= p + q + 2 – 1

= p + q + 1

p + q + 1 = 0 …. (1)

If x + 2 is a factor of f(x), then

F(-2) = 0

F(-2) = 2(-2)3 + p(-2)2 – (-2) + q

= 2(-8) + p(4) – (-2) + q

= -16 + 4p + 2 + q

= 4p + q -14

4p + q -14 = 0 … (2)

Solving simultaneously…

Subtract (1) from (2)

3p – 15 = 0

3p = 15

P = 5

Substitute the value of p into (2)

q = 14 – 4p

q = 14 – 4(5)

q = 14 – 20

q = -6

Hence p = 5, q = -6

Class activity

  1. Find the remainders without performing long division when
  2. x3 +5x2 – 3x + 1 is divided by x + 1
  3. 2x3 – 4x2 + x – 3 is divided by x + 2
  4. If (3x – 1) is a factor of the polynomial f(x) = 4x3 – 4x2 – x + p, find the value of the constant p.

SUB-TOPIC 4

Zero polynomials

Given a polynomial function f(x), the value x = a such that f(a) = 0 is called the zero of the polynomial.

A zero of the function f(x) is a root of the equation f(x) = 0.

To obtain the zeros of a polynomial f(x), set f(x) = 0 and solve the equation.

Examples:

  1. Find the zeros of the following polynomials:
  2. F(x) = x2 – 7x + 12
  3. G(x) = x2 – 16
  4. 2 is a factor of 32, find the values of the constants and state the zeros of

Solution 1:

  1. Set f(x) = 0

x2 – 7x + 12 = 0

x2 – 3x – 4x + 12 = 0

x(x – 3) – 4(x – 3) = 0

(x – 3)(x – 4) = 0

x = 3 or x = 4

Hence the zeros of f(x) are 3 and 4.

  1. Set g(x) = 0

x2 – 16 = 0

(x – 4)(x + 4) = 0

x = 4 or x = -4

Hence the zeros of g(x) are 4 and -4.

Solution 2:

If 2 is a factor, it means the zeros of 2 when substituted into will give the complete value of

from 2

2

 

(x + 2) (x – 3) = 0

x = -2 or x = 3

3(3)3 2

3(-2)3 2

⇒

From equation (i),

Substitute

Hence,

3232

To get the third factor of divide by 2

 

2 32

32

  • 2

2

0 0 0

The third factor is ⇒ hence,

More generally, an nth degree polynomial will have n zeros while an equation of degree n will have n roots.

Class activity

Given that a and b are the zeros of the polynomial f(x) = x2 – x – 6 with a˃b, and that g(x) = f(x + 2), find:

  1. g(a) + g(b)
  2. g(a) – g(b)
  3. g(a) x g(b)

PRACTICE QUESTIONS

  1. Given that P(x) = ax2 + bx + 1, P(1) = 6 and P(-1) = 2, determine the values of a and b.
  2. In the identity ax2 + bx + c = x2 – 2, find a, b, c.
  3. Given that f(x) = x3 + 3x2 + 3x + 1 and g(x) = x3 +- 3x2 + 3x -1, find (a) f(x) + g(x) (b) f(x) – g(x) (c) f(x) x g(x)
  4. Given that p(x) = ax2 + bx +1, p(1) = 6 and p(-1) = 2, determine the values of a and b.
  5. Find the quotients and remainders when P(x) = 6x3 + 4x2 – x + 5 is divided by:
  6. 3x2 + 2x +1
  7. x3 – 3x + 2
  8. A polynomial is divided by x + 1. The quotient is 2x – 3 and the remainder is 3, find the polynomial.

EVALUATION

  1. The polynomial has the same remainder when divided by (x+2) and (x-1). Find the value of constant q. (A) -11 (B) -9 (c) -3 (D) -1
  2. The polynomial has a remainder 20 when divided by (x – 2). Find the value of constant P. (A) 8 (B) 6 (C) -6 (D) -8
  3. Find the remainder when is divided by x-2 (A) 28 (B) – 28 (C) – 56 (D) 56
  4. The expression leaves a remainder of 6 when divided by x-k. The positive value of k is (A) 1 (B) 2 (C) 3 (D) 4
  5. Given f(x) = ax3 + 2x2 + bx + c, and f(0) = f(1) = 4 and f(-2) = 8, find the values of a, b and c.
  6. When is divided by (x-1), (x+2) and (x-3) the remainders are -14, 16 and 36 respectively. Find the values of the constants p, q and r and hence determine the quotient and remainder when f(x) is divided by x-4.

WEEK TWO

TOPIC: POLYNOMIALS 2

SUB-TOPICS:

  1. Roots of cubic equation: Sum of roots; Sum of products of two Roots; Products of roots.
  2. Graphs of polynomial function.

SUB-TOPIC 1

The general cubic equation takes the form: 3 (because if it becomes a quadratic equation).

3 …(1)

Dividing through by a,

3 …(2)

Let the roots of this equation 2 be. Then 3

= 0 …(3)

2

2– 2

Collect like terms

…(4)

By comparing coefficients of equations (2) and (4)

  1. The sum of roots
  2. The sum of product of roots
  3. The product of roots

Example 1:

The roots of a cubic equation are such that obtain the equation the roots of which are 2, 2, 2.

Solution:

The three ways of obtaining 2, 2 and 2 include.

(a). Expanding 2

22+2

= 2 22

2+22

⇒ 22+22

2+22 2

2+222

(b). Expanding 2 we have:

2222222

2

(c). 22222

The required equation is 322+2)222+2(22)]2 = 0

3 2

32

Example 2:

One of the roots of the cubic equation.

Find the:

  1. Sum of the two other roots;
  2. Product of the two other roots.

Hence or otherwise, find the other two roots.

Solution

Let α, β and γ be the roots of the equation such that γ = 5, then

Given that

a = 1, b = -9, c = 23, d = 15

hence:

… (1)

Also,

= 15

15

= 3 … (2)

From equation (1) we have

Substituting 4 – for

= 3

If α =1, then β =3 and

If α = 3, then β = 1

Hence:

α=1; β=3,γ=5.

Class activity

  1. The equation 32 has roots Find the equation whose roots are 3,3,3.
  2. Write down the cubic equation with solutions such that and

SUB-TOPIC 2

Graphs of polynomial function

The shape of a polynomial graph depends on the degree of that polynomial.

Polynomials of degree one

The straight line is the graphical representation of polynomials of degree one. The coefficient of x gives us a measure of the gradient or slope of the line.

If a ˃ 0, the straight line rises as y increases when x also increases. If a ˂ 0, the straight line falls as y decreases when x increases. If the graphs below, the points A and B on the straight line are called the x and y intercepts respectively.

y

Y = ax – b

a ˂ 0

B

y

Y = ax +b

a ˃ 0

B

A

x

A

x

Determining x- and y-intercept

To find the x-intercept, put y = 0 and solve for x in the equation. The x-intercept is identified as the zero of the corresponding polynomial.

To find the y-intercept, put x = 0 and solve for y.

From the knowledge of the intercepts, one can easily sketch the graph of a polynomial of degree 1.

Polynomial of degree two

The parabola is the graphical representation of polynomials of degree two. It has two shapes which depends on whether the coefficient of x2 is positive or negative.

Determining x- and y-intercept

To find the x intercept, put y = 0 and solve for x. the values of x for which y = 0 are the zeros of the polynomial.

To find y-intercept, put x = 0.

Turning points

The lowest point A on the curve in graph 1 is a turning point and it is called Minimum point.

The highest point B on the curve in graph 2 is also a turning point and it is called Maximum point.

Polynomials of degree three

The curve of polynomials of degree 3 is usually called cubical parabola and it has two shapes depending on whether a ˃ 0 or a ˂ 0.

Examples:

  1. Sketch y = 2x -1 by first finding the slopes and intercepts on the axes
  2. Sketch y = x2 +2x – 3 showing the intercepts and turning points.
  3. Sketch the curve represented by y = 12 + 4x -3x2 – x3.

Class activity

  1. Show that (2x-1) is a factor of the polynomial f(x) = 8x3 – 8x2 + 1 and find the quadratic factor.
  2. Sketch the graphs of the following: (i) y = -3x + 2. (ii) y = 8 – 2x – x2. (iii) y = x3 + 2x2 – 5x – 6.

PRACTICE QUESTIONS

  1. The expression px2 + qx +6 is divisible by x-3, and has a remainder of 20 when it is divided by x + 1. Find the values of p and q.
  2. When the polynomial f(x) = px3 + qx + r (where p, q and r are constants) is divided by (x + 3) and (x – 2), the remainders are -12 in each case. If (x + 1) is a factor of f(x), find: (i) f(x); (ii) the zeros of f(x).
  3. Given that x – 2 is a factor of 2x3 – x2 – 8x +4, find the other two factors.
  4. The equation 32 has roots Find the equation the roots of which are 3,3,3.
  5. Factorise completely 4x3 – 8x2y – 9xy2 + 18y3.

EVALUATION

  1. If the polynomial x3 + px2 + qx – 6 has a factor (x – 1) and leaves a remainder of -24 when divided by (x + 1):
  2. find the constants p and q.
  3. factorise the polynomial completely and find its zeros.
  4. Factorise 432 and 32 completely.
  5. Write down the cubic equation with solutions such that and
  6. The remainders when f(x) = x3 + ax2 + bx + c is divided by (x – 1), (x + 2) and (x – 2) are respectively 2, -1 and 15, find the quotient and remainder when f(x) is divided by (x + 1).
  7. If the polynomial f(x) = ax2 + 13x = b and g(x) = 4x2 + px + q are divided by x – 1, the remainders are 12 and 16 respectively. It they are divided by x – 2, the remainders are 40 and 20 respectively. Find the values of the constant a, b, p and q and hence determine the values of x for which f(x) = g(x).

WEEK THREE

TOPIC: PERMUTATIONS

SUB-TOPICS:

  1. Permutation (arrangement).
  2. Cyclic permutation.
  3. Arrangement of identical objects.
  4. Arrangement in which repetitions are allowed.

SUB-TOPIC 1

Permutation (Arrangement)

Suppose we are interested in the different arrangement of two people in a line. If these two people are labelled a and b then the problem is the same as finding the different arrangement of the letters a and b.

The number of arrangement will be two i.e. ab and ba.

Suppose there are three people in the line labelled a, b, and c. finding the different arrangements will be the same as finding the arrangements of the letters a, b and c.

c

b

a

c

b

b

a

c

a

b

a

c

c

a

b

bac

bca

abc

cab

cba

acb

From the above, we see there are six different arrangements.

We can get the number of different arrangement of an arbitrary n terms.

NNo. of different arrangementsFormula
111
222 x 1
363 x 2 x 1
4244 x 3 x 2 x 1
51205 x 4 x 3 x 2 x 1

Based on the above pattern, the number of different arrangement of n objects will be:

. This product can be written as n for short. n is read n-factorial.

The factorial of a positive integer is the product of all integers less than or equal to that given number.

I have four balls of different colours: Blue (B), Green (G), Red (R) and Yellow (Y). If I pick three of the balls, the following are the possible results of picking in order:

BGR BRG BGY BYG BRY BYR

GBR GRB GBY GYB GRY GYR

RBG RGB RBY RYB RGY RYG

YBG YGB YBR YRB YGR YRG

TOTAL = 24

Each of these arrangement is called a permutation. For the above, we obtained 24 permutations of four (4) colours taking three at a time. The way this is done is as follows:

The 1st ball could be any of the four balls available;

The 2nd ball could be any of the three colours remaining;

The 3rd ball could be any of the two colours remaining.

Thus, we have. This is called the arrangement of 4 balls taking 3 at a time.

If we have five colours to arrange, taking 5 at a time, we will obtain permutations.

We apply the basic counting principle that: ‘‘If an activity ‘A’ can be performed in m and another activity ‘B’ can be performed in , then, the two activities can be performed one after the other in

Permutation can therefore be defined as each of several possible ways in which a set or number of things can be ordered or arranged. It is also all possible arrangement of a collection of things where the order is important.

Suppose we are only interested in the number of ways, the first and second positions can be taken by 4 people in a race, assuming there is no tie.

The first position can be taken in four ways by any of the four athletes. The second position can be taken in 3 ways by any of the remaining 3 athletes.

So, the number of ways the first and the second positions can be taken by four people in a race is. This arrangement is called permutation of 4 people taking 2 at a time and is denoted by

 

Also, the number of permutations of 8 objects taking 3 at a time is denoted by

 

 

Thus, the general formula is

 

This is the permutation of n objects taking r at a time.

Examples

  1. Evaluate the following: (a) 7! (b) 0! (c) 1!
  2. Simplify
  3. Evaluate each of the following: (a) (b)
  4. Find the number of ways of arranging the letters of the word, EIGHT.

Solution:

  1. (a)

(b) 0! = 1

(c) 1! = 1

2.

  1. (a)

(b)

  1. EIGHT has five different numbers. Hence, the number of permutation is

 

Class activity

  1. Evaluate each of the following: a) b)
  2. In how many ways can five bulbs of different colours be arranged in five socket in a row?
  3. In how many ways can the letters of the word ENGLISH be arranged?

SUB-TOPIC 2

Cyclic permutation

In cyclic permutation, we are concerned about arrangement of this about a circular object.

If the letters A, B, C, D are arranged in that order in a circle, and then A is moved to B’s position and B to C’s position, C to D’s position, and D to A’s position, we obtain the same arrangement. i.e,

A

D

C

B

D

C

B

A

 

To obtain different arrangement, we fix one of the letters and arrange the remaining three in the remaining spaces. This gives 3! arrangements of the 4 letters.

A

D

C

B

A

C

D

B

A

D

B

C

A

C

B

D

A

B

D

C

A

B

C

D

In general, the number of ways of arranging ‘n’ objects in a circle is given by:

No. of ways = 1 x (n-1)!

When beads are threaded in a ‘ring’ the clockwise and the anticlockwise arrangements are not distinguishable and the ring can be turn over.

Thus, the number of distinct arrangements of ‘n’ objects round a circular ring which can be turned over is:

Examples:

  1. In how many ways can 8 boys be arranged at a round table?

Solution:

Total number of arrangements = 1 x (8-1)! = 7! = 5040

  1. Seven beads of different colours are threaded in a ring. How many different arrangement is possible?

Solution:

No. of arrangements =

Class activity

  1. In how many ways can eight boys be arranged around a circular table?
  2. A family of seven is to be seated round a table. In how many ways can this be done if the father and the mother are to sit together?
  3. In how many ways can eight men be seated at a round table if two particular men refused to sit together?

SUB-TOPIC 3

Arrangements of identical objects

Consider the arrangement of 8 bulbs (4 red, 3 blue and 1 yellow) in a row, there are 8! Possible permutations (arrangements). Out of these permutations, 4! Permutations involving changes in position of the red bulbs are not distinguishable and the 3! Permutations of the blue bulbs are also not distinguishable.

Thus, the number distinct permutations of the 8 bulbs (4 red, 3 blue and 1 yellow) =

In general, the number of distinct permutations of the ‘n’ objects containing p of one type of q of a second type and r of a third type is given by:

Example

In how many ways can the letters of the following words be arranged ?

  1. ABAKALIKI
  2. MATHEMATICS

SOLUTION

a) ABAKALIKI 9 letters; letter A appears 3 times, K twice, I twice, B&L once each.

No. of permutation =

b) MATHEMATICS = 11letters ( 2 M’s, 2 T’s, 2 A’s, & others once)

 

Class activity

  1. Find the number of ways the letters of the words FURTHER can be permuted.
  2. In how many ways can the letters of the word STRANGE be arranged so that the vowels occupy only the odd places?
  3. Find the number of arrangement in the letters of the word CONGRATULATIONS if the letter A must be placed next to each other. (Leave your answer in terms of factorial).

SUB-TOPIC 4

Arrangements in which repetitions are allowed

Examples

  1. How many numbers greater than 600 can be formed from the digits 2, 3, 4, 5, 6, 7?
  2. How many four-digits even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 if
  3. No repetitions are allowed
  4. Repetitions are allowed.

SOLUTION

(a) Without repetition: Numbers greater than 600 that can be formed from the option of 6 digits 2, 3, 4, 5, 6, 7 can be a 3-digit number, 4-digit number, 5-digit number or 6-digit number.

For a 3-digit number to be greater than 600, the first digit must be any of the 2 digits (6 or 7)

The 2nd digit, any of the remaining 5 digits ( i.e, after choosing one digit from 6 or 7)

The 3rd digit, any of the remaining 4 digits.

Therefore,

For 3-digits numbers we have

For 4-digits numbers, we have

For 5-digits numbers, we have

For 6-digits numbers, we have

Total = 40 + 360 + 720 + 720 = 1840

With repetition: the numbers of digits greater than 600 that would be formed with 2, 3, 4, 5, 6 and 7 will be:

For 3 digits numbers, there will be

For 4 digits numbers,

For 5 digits numbers,

For 6 digits numbers,

Therefore, total numbers that can be formed = 72+1296+7776+48656 = 57,800

(b) For an even number, the last digit must be 2 or 4 or 6.

Without repetition: The last digit can be any of the 3 digits.

The 1st digit can be any of the remaining 5 digits

The 2nd digit can be any of the remaining 4 digits

The 3rd digit can be any of the remaining 3 digits.

Therefore number required =

With repetition: the number required =

Class activity

  1. How many 4-digits numbers can be formed from the digits 0, 1, 2, 3, …9

If: (a) repetitions are allowed (b) the last digit must not be zero and repetitions are not allowed.

  1. How many 4-digit odd numbers can be formed with the digits 1, 2, 3 and 4 if: (a) repetition is allowed (b) repetition is not allowed.
  2. How many numbers less than 3000 can be formed from the digits 1, 2, 3, 8 and 9, if no digit is used more than once?

PRACTICE QUESTIONS

  1. Simplify. (a) 24 (b) 42 (c) 72 (d) 27
  2. Five students are lined up in a row. How many arrangement could be made if the position of the last boy remains unchanged? (a) 120 (b) 12 (c) 21 (d) 24
  3. Find the number of ways in which the letters of the word STATISTICS could be arranged. (a) 15120 (b) 5120 (c) 2020 (d) 1512
  4. Seven students were late to a class. In how many ways can they occupy: (a) three available vacant seats? (b) Nine available vacant seats?
  5. In how many ways are there of arranging 3 different jobs between 5 men where any man can only do one job?

EVALUATION

  1. In how many ways can 8 people be seated on a bench if only 3 seats are available?
  2. If the mathematics department of a particular college has 5 members of staff and they are to pose for a photograph by standing in a row. How many different arrangements are possible?
  3. Five people are to have a dinner jointly. In how many ways can they sit round a table if a couple must sit together?
  4. How many numbers greater than 4000 can be formed using some or all the digits 6, 5, 4, 3 and 2 without repetition? How many of these will be even?
  5. How many words can be formed from the letters of VALEDICTORY provided the letters A, E, I, O, Y are not to be separated at all?

WEEK FOUR

TOPIC: COMBINATION

SUB-TOPICS:

  1. Combination (selection).
  2. Conditional arrangements and selection.
  3. Probability problems involving arrangement and selection.

SUB-TOPIC 1

Combination (Selection)

In many situations, we make selection without regard to the order. If a committee of 4 members is to be formed from 7 members of staff of DLHS, the order in which the numbers of the given committee are selected is not important.

Combination is therefore a way of selecting items from a collection such that (unlike permutation) the order of selection does not matter.

In selecting three colours from 5 colours: (B, G, R, W, Y), BGR, BRG, GBR, RBG, RGB, are counted as 6 different arrangements (permutations), although they consist of the same 3 colours. The 6 permutations thus represent one combination. Thus, each combination of three objects yields 3! permutation.

Now, the number of the permutations of 5 colours taking 3 at a time, i.e,

The number of combinations of 5 colours taking 3 at a time, i.e,

In general,

Examples

  1. Out of the five science club members of a school, A, B, C, D and E, just three are to be chosen to represent the school in an exhibition. In how many ways can this be done?
  2. In how many ways can a committee of 3 chemistry teachers and 5 mathematics teachers be formed from 6 chemistry teachers and 10 mathematics teachers?

Solution

  1. The three representatives can be selected in
  2. The chemistry teacher can be selected in 6C3 ways and the mathematics teacher can be selected in 10C5 ways.

Total number of ways = 6C3 x 10C5

=

=

=

=

Class activity

  1. In how many ways can a disciplinary committee of 3 be formed from 10 members of staff of a college?
  2. A man travels from Nigeria to Ghana by air and from Ghana to Liberia by ship. He returns by the same means. He has 6 airlines and 4 shipping lines to choose from. In how many ways can he make his journey without using the same airline or shipping line twice?

SUB-TOPIC 2

Conditional arrangement and selection

When restrictions are placed on arrangements or selection, then, the permutation or combination is said to be conditional.

Example 1:

Find the number of ways 6 people can be seated in a round table if two particular friends must sit next to each other.

Solution:

If two people must seat next to each other, the number of ways these friends can sit is 2!

Therefore, the number of ways six people can sit in a round table with two friends that must be together is

Example 2:

A committee of 4 people is to be chosen from 5 married couples. Find how many ways the committee can be chosen if: (i) everyone is equally eligible; (ii) the committee should include at least one woman.

Solution:

i) 5 married couples includes 5 men and 5 women. Since everyone is equally eligible, then, the possible ways of selecting 4 people for the committee are:

4 men and 0 women or

3 men and 1 women or

2 men and 2 women or

1 man and 3 women or

0 men and 4 women.

i.e,

 

 

ii) If at least one woman must be in the committee, then, the possible ways of selecting 4 members of the committee from the couples (5men & 5women) are:

3 men and 1 woman or 2 men and 2 women or 1 man and 3 women or 4 women

i.e,

 

Class activity

  1. An excursion group of 4 is to be drawn from among 5 boys and 6 girls. Find the number of ways of choosing the excursion group if the group:
  2. is to be made up of an equal number of boys and girls;
  3. is to be either all boys or all girls;
  4. has no restrictions on its composition.
  5. A candidate is expected to attempt 12 out of 15 questions. In how many ways can this be done if:
  6. the candidate is to attempt any 12 question;
  7. the first 8 questions are compulsory;
  8. a question is outside the syllabus and hence cannot be completed.

SUB-TOPICS 3

Probability problems involving arrangement and selection

Example 1:

A box contains 10 red, 3 blue and 7 black balls. If three balls are drawn at random, what is the probability that: (a) all 3 are red, (b) all 3 are blue, (c) one of each colour is drawn?

Solution:

Number of ways of selecting any 3 balls from 20 balls = number of element in the sample space.

Number of ways of selecting 3 red balls out of 10 =

(a) P(all the 3 balls are red) =

(b) P(all 3 are blue) =

(c) P(1 red, 1 blue & 1 black) =

Example 2:

Three-digit numbers are formed from the digits 1, 2, 4, 5 and 6. If repetition is not allowed and a number is picked at random, find the probability that it is a multiple of 5 or an odd number.

Solution:

Since probability is involved, we find the sample space for 3-digit numbers formed from 5-digits i.e 5P3.

n(sample space) =

  1. To get multiple of 5, the last digit must be 5. That implies that

The 1st digit can be any of the remaining 4 digits.

The 2nd digit can be any of the remaining 3 digits

No. of ways = 1 x 4 x 3 = 12 ways.

Pr(multiple of 5) =

  1. To get odd number, the last digit will be any of 1 and 5

The first digit can be any of the remaining four

The second digit will be any of the remaining three.

No. of ways = 2 x 4 x 3 = 24ways.

Pr(odd number) =

Hence, pr(multiple of 5 or odd number) =

=

=

Class activity

  1. How many committee of size 5 consisting of three men and 2 women can be selected from 8 men and 6 women if a certain man must not be in the committee?
  2. A bag contains 5 white, 2 black and 3 green balls. If three ball are drawn at random, find the probability that:
  1. All three are green
  2. All three are white
  3. 2 are white and 1 is black
  4. At least, one is black
  5. 1 of each colour is drawn.

PRACTICE QUESTION

  1. How many committee of size 5 consisting of 3 men and 2 women can be selected from 8 men and 6 women if a certain man must not be in the committee? (a) 315 (b) 525 (c) 840 (d) 1287
  2. In how many ways can 9 bulbs be selected from 4 red, 5 green and 6 yellow bulbs if 3 of each colour are to be selected? (a) 800 (b) 120 (c) 40 (d) 27
  3. The number of ways of arranging 9 men and 8 women in a row, when the women occupy the even places is — (a) (b) (c)!8! (d)
  4. A panel consists of 5 men and 4 women. What is the probability of 4 men and 2 women?

(a) (b) (c) (d) .

5. Five digit numbers are formed from digits 4, 5, 6, 7 & 8

  1. How many of such numbers can be formed if repetition of digit is (i) allowed (ii) not allowed?
  2. How many of the numbers are odd if repetition of digits is not allowed?

EVALUATION

  1. Find the number of ways of arranging 9 men and 8 women in a row, if the women occupy the even places.
  2. If find the value of n.
  3. A panel of 5 jurist is to be chosen from a group of 6men and 7 women. Find the number of different panels that could be formed if: (a) a particular man must serve on the panel (b) there is no restriction.
  4. A business man intends to give a dinner party for 6 of his 10 friends. If 2 of them will not attend the party together, in hoe many ways can he select his guests?
  5. A family of 7 is to be seated round a table. In how many ways can this be done, if the father and the mother are to sit together?
  6. Four delegates are to be chosen from 8 members of staff of a college. If 2 of them are senior members of staff, how many different delegations are possible if: (i) only one of the senior members of staff must be in the delegation? (b) the two senior member of staff must be included?

WEEK FIVE

TOPIC: BINOMIAL EXPANSION 1

SUB-TOPICS:

  1. Pascal triangle.
  2. Binomial expansion of (a+b)n , where n can be positive integer, negative integer or fractional value.

SUB-TOPIC ONE

The Pascal’s triangle is a format for getting the coefficients of expansions. It applies to binomial and binomial formed from a reduced polynomial.

Consider the expansion of each of the following:

(a+b)0

(a+b)1

(a+b)2

(a+b)3

(a+b)4

(a+b)5

By ordinary expansion of algebraic terms, we have:

(a+b)0 = 1

(a+b)1 = 1a + 1b

(a+b)2 = 1a2+2ab+1b2

(a+b)3 = 1a3+3a2b+3ab2+1b3

(a+b)4 = 1a4+4a3b+6a2b2+4ab3+1b4

(a+b)5 = 1a5+5a4b+10a3b2+10a2b3+5ab4+1b5

Consider the array of coefficients of a and b. We can display it as follows:

 

n (power)

 

1 0

1 1 1

Row 1 2 1 2

1 3 3 1 3

1 4 6 4 1 4

1 5 10 10 5 1 5

We call the array of coefficients displayed above Pascal triangle named after the celebrated French Mathematician and Physicists Blaise Pascal (1623-1662) noted for his essay on conic section in 1640 and first invention of calculating machine in 1642.

Note the following:

  • A general binomial is of the form (a+b)n .
  • There are n + 1 terms.
  • The expansion is homogenous i.e.the sum of the powers of a and b in each term of the expansion is n.
  • As the power of a descends (starting from n till it reaches 0), the power of b ascends (starting from 0 till it reaches n) and vice versa.

Examples:

  1. Using Pascal’s triangle, expand

Solution:

Let n=4

1 4 6 4 1

Understand that the index (power) of descends as that of ascends.

 

  1. Use Pascal’s triangle to obtain the value of (1.025)4, correct to three decimal places.

Class activity

  1. .

SUB-TOPIC 2

Binomial Expansion of (a+b)n

Binomial Theorem

ncr

ncr =

ncr = ncr

The Binomial Theorem for a positive Integral Index

If a and b are any numbers and n is a positive integer, then

nC0an+nC1an-1b+nC2an-2b2+…+nCran-rbr+…nCnbn

Note:

  1. The number of terms in the expansion is n+1. That is one more than the index of binomial.
  2. The (r+1) term in the expansion of the binomial is called the general term and denoted by Tr+1 =nCran-rbr

The Binomial Theorem for Negative and Fraction

When n is not a positive integer, the expansion becomes

 

Provided a is numerically less than unit. That is -1<< 1. This means that the various coefficients cannot be expressed as nC0, nC1, nC2 etc because they have no meaning when n is not a positive integer.

Again, the theorem can be applied only when the first term of the binomial is unity. If not, the binomial must first be reduced to this form. For example, to expand (t+a)n has to be put in this form .

Examples:

4C0 + 4C1 + 4C2 + 4C3 + 4C4

=

  1. Expand to five terms.

Solution:

Recall:

n = -3,

  1. Expand to four terms.

Solution:

Note the first term of the binomial is not unity. Let’s reduce to the form

n= ½,

Class activity

  1. Determine the coefficient of from the expansion of .
  2. Expand to five terms.
  3. Expand to three terms.

PRACTICE QUESTIONS

  1. Using the binomial theorem, expand (1 + 2x)5, simplifying all the terms. Hence calculate the value of (1.02)5 correct to six significant figures.
  2. If the first three terms of the expansion of the expansion of (1+px)5 in ascending powers of x are 1+20x+160x2 find the values of n and p.
  3. (a) Write down the binomial expansion of (1+y)8, simplifying all the terms.

(b) Using the substitution y = x – x2 in (a), deduce the expansion of (1+ x – x2)8 in ascending powers of x as far as the term in x4.

(c) Find, by inspection, a value of x such that 1+ x – x2 = 1.09. Hence, evaluate (1.09)8 correct to three decimal places.

  1. Write down the binomial expansion of , simplifying all its coefficients.
  2. Obtain the first five terms of the expansion of

ASSIGNMENT

        1. The _______ triangle is a format for getting the coefficient of expansions. (a) Binomial (b) right-angled (c) Pascal’s (d) array.
        2. Pascal’s triangle was named after the French Mathematician and Physicists called_________ (a) Newton (b) Blaise Pascal (c) Cramer (d) Laplace
        3. The general binomial is of the form ______ (a) (a+b)n (b) (a+b)r (c) (a+b)! (d) aCb
        4. The (r+1) term in the expansion of the binomial is called the ________ term. (a) Simple (b) Difficult (c) Binomial (d) General.
        5. Expand
        6. Expand.

WEEK SIX

TOPIC: BINOMIAL EXPANSION 2

SUB-TOPICS:

  1. Finding nth term.
  2. Application of binomial expansion

SUB-TOPIC 1

Finding the nth term

The binomial expansion of is given as:

=

This expansion is true for any natural number value of n, large or small, but when n is a large natural number then is small. Indeed, the larger the value of n, the closer becomes to zero. The n notation for this,

 

the limit of as n→ is 0.

Also, as n→, the closer the expansion above becomes to the sum of terms:

Here the ellipsis (…) at the end of the expansion means that the expansion never ends, that is it has infinite number of terms.

Now we can use the sigma notation and write:

 

Notice the symbol for infinity (∞) at the top of the sigma, this denotes the fact that the sum is a sum of an infinite number of terms.

Examples:

  1. How many terms will there be in the expansion of. Find the fourth term.

Solution:

n=11, x=2x and

  1. Number of terms = n+1 = 11+1 =12
  2. From the formula,

11-3r = 2

-3r =2-11

-3r = -9; 3r = 9 ; hence r=3 (i.e. the 4th term)

 

  1. Find the term x2 and then term independent of x in the expansion of find the fifth term:

Solution:

  1. The general term is

When it is x2, then 12-2r = 2

r=5 (i.e. the sixth term)

Hence,

=

=

=

  1. If the term is to be independent of x, then 12-2r = 0

r =6 (i.e. the 7th term)

= = =

The independent term of x is

Class activity

  1. Find the term independent of x in the expansion
  2. How many terms will there be in the expansion?

SUB-TOPIC 2

Application of Binomial expansion in approximation

If x is very small, we can take 1 + nx as an approximation of is taken as 1-nx.

Examples:

Find the linear approximation of the following:

  1. (1.01)5 (b) (1.02)4 (c) (1.05)3 (d) (0.98)5

Solution: (a) (1.01)5

Expansion of (1+x)n = 1+nx= (1+0.01)5=1+5×0.01= 1+0.050= 1.050

  1. (1.02)4

Solution: Expansion of (1+x)n = 1+nx

(1.02)4 = (1+0.02)4≈1+4×0.02

= 1+0.080= 1.080

(c) (1.05)3; Expansion of (1+x)n = 1+nx

(1.05)3 = (1+0.05)4≈1+3×0.05= 1+0.15= 1.150

(d) (0.98)5

Solution: Expansion of 1+xn = 1+nx

(0.98)5 = (1-0.02)5≈1-5×0.02= 1-0.1= 0.09

EVALUATION

Find the linear approximation of the following:

  1. (1.002)15 (ii) (0.99)8 (iii) (2.004)7

PRACTICE QUESTIONS

  1. How many terms will there be in the expansion of
  2. Find the terms in the expansion of . Find the fift term.
  3. Expand and hence evaluate (0.98)7 correct to three decimal places.
  4. Find the linear approximation of the following (a) (2.004)7 (b) (1.003)11 (c) (d) .
  5. Using the binomial theorem, write down and simplify the first seven terms of the expansion of (1+2x)10 in ascending powers of x. use your expansion to show that 1.210˃6.19.

WEEKEND ACTIVITY:

  1. Find the number of terms in the expansion of . Find the third term.
  2. Find the term x2 and independent of x in the expansion .
  3. Find the coefficient of x-4 in the expansion.
  4. Determine the coefficient of from the expansion of .
  5. Expand to four terms.

WEEK SEVEN

MID TERM BREAK.

WEEK EIGHT

TOPIC: ROOTS OF QUADRATIC EQUATION 1

SUB-TOPICS:

  1. Quadratic equation (completing the square and formula method).
  2. Sum and product of roots of quadratic equation.
  3. Finding quadratic equation given sum and product of roots, x2 – (sum of roots) + (product) = 0.
  4. Condition for quadratic equation to have: (i) Equal roots (b2 = 4ac) (ii) Real roots b2 > 4ac (iii) No roots b2 < 4ac

SUB-TOPIC 1

Quadratic equation (completing the square and formula method).

A quadratic equation (trinomial) in one variable is a three termed equation in which the highest power of the variable is two. The general quadratic equation in variable x is of the form ax2 + bx + c = 0 where a ≠ 0.

In general, a quadratic equation has two solutions which may or may not b+e equal.

There are four major methods of solving quadratic equations. They are:

  • factorisation method;
  • completing the square method;
  • formula method;
  • graphical method.

Factorisation

Examples

Solve the following by factorisation

Solution

(a) Product of 1st and 3rd term = -8x2

Factor of -8x that will sum to -2x (the middle term) = -4x and +2x

Replacing the middle term with the two factors, we have:

(b) Product = -120x2

Sum = 14x

Factors = -6x and +20x

Completing the square method

Given ax2 + bx + c = 0, a ≠ 0

Write the equation as ax2 + bx = -c

Divide each term by the coefficient of x2. Find half of the new coefficient of x, square it and add this to both sides of the equality sign. This makes the expression on the LHS a perfect square.

Take the square root of both sides and determine the values of x to the desired accuracy.

The above is the process of solving quadratic equation using the method of completing the square.

Quadratic formula

The formula derived above is the quadratic formula.

Example:

Use the quadratic formula to solve

Solution:

a = 2; b = 7; c = -15

Class activity

  1. Use the three methods discussed above to solve the following quadratic equations:
  2. 2x2 – 17x – 9 = 0
  3. 3x2 + 10x – 12 = 0
  4. 8x2 + 34x + 21 = 0

SUB-TOPIC 2

Sum and product of roots of quadratic equation

Recall the completing the square method and general formula ‘ of the general equation

In this case the two roots of the equation are = or

It is possible to represent the two roots by then we have

 

 

 

 

 

 

Product of roots

 

From the above, if we have

We can rewrite this as .

Also, if α and β are the roots of the quadratic equation, then;

 

 

 

Therefore,

and

Given a quadratic equation, we can find the sum and product of the roots using the above information.

Example 1: if the roots of the equation are

Find

Solution: In the equation

 

The roots of the equation are

Then (a)

(b)

(c)

From (a)

 

(d)

 

 

24

= 40

Example 2:- if the roots of the equations

are (c) α2 – β2

Solution :- (i) The equation is

 

 

 

 

has no expression, then

 

 

 

But we want

 

 

 

 

 

This cannot be determined

The equation is

 

 

 

Following example I (c) we can express this as

(

 

 

 

Example 3:- Express the following in terms of

 

Solution:-

  1. .

 

 

.

 

 

Class activity

  1. are the roots of the equation

Find

 

 

  1. Given , prove that

 

 

SUB-TOPIC 3

Find quadratic equation given sum and product of roots

We have been able to establish that given α and β as the roots of a quadratic equation where a, b and c are constant and, then,

 

Also, and

Therefore, given the roots of a quadratic equation, the equation can be gotten this way:

Example 1:-

Given that the roots of equation are 3 and 7, find the equation.

Solution:

Roots of the equation are 3 and 7

Sum of roots

Product of roots (

 

 

Example 2:- Find the equation whose roots are -8 and 2.

Solution: Roots of the equation are -8, 2

 

 

The equation is

 

Example 3: If the roots of the equation are, find the equation.

Solution: Roots are

 

 

 

 

Multiply through by 12

 

Class activity

  1. Construct and simplify equations whose roots are given below:

 

  1. , write out the equation whose roots are . Find the values of from your equation.

SUB-TOPIC 4

Condition for quadratic equation to have:

 

In this section, we want to see some properties of roots. We can determine the type of roots that a particular equation will have.

The quadratic equation of

The part of the roots under square root sign is called the discriminant (i.e.) of the roots of the quadratic equation. This is because it can be used to determine the nature of the roots.

 

This means that whatever is contained inside the square root.

 

The quadratic is said to have coincident roots. This happens when the quadratic equation is a perfect square.

Example 1:- Consider the equation

SOLUTION: By factorization

 

 

 

 

(ii) Real roots: When b2 is greater than 4ac then the value under the roots sign is a positive number that is . The square root of such number will have two real values, one positive and other negative. The equation is said to have two distinct real roots.

Example 2: Solve the equation

 

,

 

 

 

(iii) Imaginary roots: When b2 is less than 4ac then the value under the root sign is a negative number i.e This shows that the root is not a real number. We say that the equation has imaginary roots.

Example 3: Find the roots of the equation

Solution: a=1 , b=2 c=5

Then

 

The roots are imaginary roots

Example 4:- Without solving the equation determine whether the equation has two different roots, coincident roots or imaginary roots.

Solution:- Rewrite the equation

 

Then a=5, b=-5, c=1

Then b2 = 25

 

 

Since

Then the equation has two real different roots.

Example 5:- Determine the nature of the roots equations without solving them.

 

Solution :

Rewrite the equation

 

To find the nature of the roots

 

 

 

 

The equation has two distinct real roots.

 

 

 

 

The equation has no real roots.

Class activity

Determine the nature of the roots of the following equations without necessarily solving them:

 

  1. Two real and different roots?
  2. Coincident and real roots
  3. No roots

PRACTICE QUESTIONS

        1. Given that are the roots of an equation such that the equation..
        2. If the equation has coincidental roots, find the value of P. (a)
        3. If
        4. If ( where p and q are constant, find the possible values of q.
        5. The roots of the equation are α and β, with α greater than β. Find the values of: (i) α – β (ii) α2 – β2.
        6. If the roots of the equation 4 Find the equation whose roots are

ASSIGNMENT

The roots of the equation are where find the value of k.

Find without necessarily solving the equation, the nature of the roots of the equation 3 . The equation (a) distinct two roots. (b) has no roots (c) coincident real roots (d) none of the above

If are the roots of the equation find (a)

Form the equation whose roots are -4 and 9.

The roots of the equation are, where m is a constant. If find the value of m.

If the sum of the squares of the roots of the equation is 1, show that

WEEK NINE

TOPIC: ROOTS OF QUADRATIC EQUATIONS 2

SUB-TOPICS:

  1. Quadratic functions (Simultaneous Equations: One linear, One quadratic)
  2. Solution of problems on roots of quadratic equation.
  3. Maximum and minimum values.

SUB-TOPIC 1

Quadratic functions (Simultaneous Equations: One linear, one quadratic)

We have discussed different ways but we need to mention that graphical solution is very important aspect of solving quadratic equations. This is because with graphical solution a lot of other problems can be solved.

The graph of the quadratic equation called parabola. Some call it cup or cap. The quadratic expression is equated to y and it is called a quadratic function. The example below show the graphical solution of quadratic function.

Example 1: Solve graphically, the equation 2

Solution:

Draw the table of values for the equation 2

01234
22712303122748
3210-1-2-3-4
-2-2-2-2-2-2-2-2
28124-2082242

Choose a convient scale, on and, on let represents and on represents

From the graph we find the point here the curve intersects

The graph is also useful to determine the minimum value of the minimum value of we have minimum point when and maximum point when

Simultaneous Equations

When solving simultaneous equation (you are already used to solving it graphically). In situation where one equation is linear and the second is quadratic, it can be solved by substitution as well as solving graphically.

In graphical solution of one linear-one quadratic simultaneous equation, there are three possible relationships between the straight line (linear) and the parabola (quadratic). They are:

  • Line intersecting with curve
  • Line touching curve at a point (tangent)
  • Line not intersecting the curve.

Example 2: Solve the simultaneous equations: 22

Solution: By substitution:

22

22

Since, then

Hence, adding (iii) together we get

From (ii) .

Example 3: Given the simultaneous equations:

2 and

Show on the graph the points of interest. Hence write out the values of .

Solution:

2

 

Table of values for 2 and

2

2362516941014

From the above, there was no intersect of the curve and the straight line. The solutions to the two equations cannot be determined because there is no point of intersection.

The points of intersection give the solution.

Example 3: On the same axes, plot the graph of y = 2x2-5 + 4 and y = 2x + 3. Hence find the points of intersection of the two graphs.

Solution:

Prepare the table of values for the functions given above.

y = 2x2-5 + 4

X-4-3-2-10123456
2x232188202818325072
-5x20151050-5-10-15-2025-30
+4+4+4+4+4+4+4+4+4+4+4+4
Y563722114127162946

Choose a convenient scale.

The points of intersections x = 0.2 and 3.3

The above example shows the case of the line intersecting with the curve.

Example 4: solve the simultaneous equation y = x2-2x + 2 and y = 4x -7. Interpret your result geometrically.

Solution:

Eliminate y to obtain: x2-2x + 2 = 4x -7 ⇒ x2 – 6x + 9 = 0

By factorisation:

(x – 3)(x – 3) = 0⇒x = 3(twice).

From y = 4x – 7 = 4(3) -7 = 5. The solution is x = 3 and y = 5.

Draw the graphs of the two equation to interpret it geometrically.

Table of values for y = x2 – 2x + 2. y = 4x – 7

x-2-101234x0234
x2410149164x081216
-2x820-2-4-6-8-7-7-7-7-7
+2+2+2+2+2+2+2+2y-7159
y145212510

 

The line y = 4x -7 intersects the curve y = x2 + 2x + 2 at only one point. Therefore, the solution to the equations is at the point x = 3 and y = 7.

Class activity

  1. Solve the simultaneous equations y = 4x – 1 and y =2x2 graphically and interpret your result geometrically.
  2. Solve for -1. Using a scale of 2cm to 1 unit on the x-axis and 2cm to represent 5 units on the y-axis.

SUB-TOPIC 2

Solutions of problems on roots of quadratic equation

Mathematics is important of life situation because of its application. You are used to problems leading to simple equations. We want to see the word problems leading to quadratic equations.

In order to solve such problems, you must take note of the following:

  1. Express the ideas involved in mathematical symbols.
  2. Write out the equation using the symbols.
  3. Solve the equation.
  4. Interprete your result.

Example 1: the product of two consecutive whole numbers is 506. Find the numbers.

Solution:

Let the numbers be x and (x + 1).

Then, x(x+1) = 506 ⇒ x2 + 1 = 506 (this is now quadratic equation)

X2 + x – 506 = 0

Solve by formula to find the values of x using the parameters below

a = 1 b = 1 c = -506

Example 2: There are two possible routes from Lagos to Ijebu Ode. One route is through Lagos/Ibadan express way which is 100km and the other is through Ikorodu-Epe covering a distance of 80km. A motorist going through express way can travel 10km per hour faster than the one going through Ikorodu and Epe and arrive Ijebu-Ode 5 minutes earlier as well. What is the time spent on the journey to Ijebu Ode by the motorist travelling through the express way?

Solution:

Let x be the speed of motorist going through Ikorodu/Epe and the speed of the one going through express way is x + 10.

Time taken by Ikorodu/Epe = 80/x.

Time taken by express way = 100/ (x + 10)

Hence, 80/x – 100/(x+10) = 1/12.

 

Form a quadratic equation from (i) above and solve it using formula and conclude.

Class activity

  1. The length of a rectangular field is 6m more than the width. If the area of the field is 72m2, find the dimensions of the field.
  2. Two consecutive odd integers are such that the sum of their reciprocals is Find the odd integers.

SUB-TOPIC 3

Maximum and Minimum values

The graph of as we have seen is a parabola. We have minimum point when and maximum point when

The maximum or minimum value (y) is

The curve is symmetrical about the line which is called the axis of symmetry.

If f(x) = 0, then,

      1. the curve cuts the horizontal axis if
      2. the curve touches the horizontal axis if
      3. the curve does not cut the horizontal axis if

Example 1:

Find the minimum value of and the corresponding the value of x for which y is a minimum.

Solution:

 

 

 

When x = -5/6, the expression in the brackets will be zero, hence the minimum is -49/12.

The corresponding value of x for which y is minimum is -5/6.

Note that x = -5/6 is the axis of symmetry of the parabola. Alternative, let the minimum value of y be ym then

 

Also the equation of the line of symmetry is

X = -b/2a = -5/6.

General evaluation:

  1. Solve the equations simultaneously and show the points of intersections

Y = 4 – 11x and y = 2x2-19

  1. Find the maximum value of y = 5 + 4x – x2 and the coordinates at the point where the curve y = 5 + 4x – x2, cuts the coordinates axes.
  2. The formula gives the sum of consecutive whole numbers. If
  3. A father got his first son at 31 years. If the product of their ages is 816. Find the ages of the father and his son.

SUBJECT: FURTHER MATHEMATICS CLASS: SSS2

SCHEME OF WORK

WEEK TOPICS

  1. Polynomials 1: (a) Definition of polynomials (b) Basic operations on polynomials. (c) Remainder and factor theorem. (d) Zeros of polynomials.
  2. Polynomials 2: (a) Roots of cubic equation: Sum of roots; Sum of products of two Roots; Products of roots. (b) Graphs of polynomial function.
  3. Permutations: (a) Permutation (arrangement) (b) Cyclic permutation (c) Arrangement of identical objects. (c) Arrangement in which repetitions are allowed.
  4. Combination: (a) Combination (selection). (b) Conditional arrangements and selection. (c) Probability problems involving arrangement and selection.
  5. Binomial Expansion 1: (a) Pascal triangle (ii) Binomial expansion of (a+b)n , where n can be positive integer, negative integer or fractional value.
  6. Binomial Expansion 2: (a) Finding nth term (b) Application of binomial expansion
  7. Mid-term break.
  8. Roots of quadratic equation 1: (a) Quadratic equation (completing the square and formula method). (b) Sum and product of roots of quadratic equation (c) Finding quadratic equation given sum and product of roots, x2 – (sum of roots) + (product) = 0 (d) Condition for quadratic equation to have: (i) Equal roots (b2 = 4ac) (ii) Real roots b2 > 4ac (iii) No roots b2 < 4ac
  9. Roots of quadratic equation 2: (a) Conditions for given line to: (i) intersect a curve, (ii) be tangent to curve, (iii) not intersect a curve (b) Solution of problems on roots of quadratic equation. (c) Maximum and minimum values.
  10. Revision.
  11. Examination.

WEEK ONE

TOPIC: POLYNOMIALS 1

SUB-TOPICS:

(a) Definition of polynomials.

(b) Basic operations on polynomials.

(c) Remainder and factor theorem.

(d) Zeros of polynomials.

SUB-TOPIC 1

Definition of Polynomials

A polynomial is a mathematical expression which is a sum of terms, each term consisting a variable or variables raised to a power and multiplied by a coefficient. A polynomial of one variable x (univariate) has the following as its general form:

anxn + an-1xn-1 + … + a2x2 + a1x + a0

where the highest power of the variable n is the degree of the polynomial; the numerical constants an, an-1, … a2, a1 are called the coefficients of the polynomial, while a0 is called the constant term.

Examples of polynomials include:

  • 3x2 – 2x + 4
  • 2x3 + 3x2 + 5x + 3

A function whose values are given by a polynomial is called a polynomial function. Eg: f(x) = 2x3 + 3x2 + 5x + 3

An equation that is obtained when we set a polynomial equal to zero is called a polynomial equation. E.g.: 2x3 + 3x2 + 5x + 3 = 0

Equality of polynomials

Two polynomials,

P(x) = anxn + an-1xn-1 + … + a2x2 + a1x + a0 and

Q(x) = bnxn + bn-1xn-1 + … + b2x2 + b1x + b0

are said to be equal if:

an = bn

an-1 = bn-1

a2 = b2

a1 = b1

a0 = b0

The value that is obtained by substituting a for x in a polynomial P(x) is denoted by P(a).

SUB-TOPIC 2

Basic operations on polynomial

Addition and Subtraction of Polynomials

Given

P(x) = anxn + an-1xn-1 + … + a2x2 + a1x + a0 and

Q(x) = bnxn + bn-1xn-1 + … + b2x2 + b1x + b0

Then,

P(x) + Q(x) = (an + bn) xn + (an-1 + bn-1) xn-1 + … + (a2 + b2) x2 + (a1 + b1) x + a0 + b0

Similarly,

P(x) + Q(x) = (an – bn) xn + (an-1 – bn-1) xn-1 + … + (a2 – b2) x2 + (a1 – b1) x + a0 – b0

Examples:

  1. Given P(x) = 5x3 – 3x2 + 4x + 7; Q(x) = 6x2 + 5x – 4; R(x) = 8x3 + 5x – 2.

Find, (a) P(x) + Q(x); (b) R(x) – P(x); (c) P(x) + 2Q(x) – 3R(x).

  1. If F(x) = 3x3 + 4x2 – 5x + 9, find: (a) F(-1). (b) F(1). (c) F(0). (d) F(3).

Solution:

  1. (a) P(x) + Q(x) = (5x3 – 3x2 + 4x + 7) + (6x2 + 5x – 4)

= 5x3 + (– 3x2 + 6x2) + (4x + 5x) + (7 + (-4))

= 5x3 + 3x2 + 9x + 3

(b) R(x) – P(x) = (8x3 + 5x – 2) – (5x3 – 3x2 + 4x + 7)

= (8x3 – 5x3) + (0 – (-3x2)) + (5x – 4x) + ((-2) -7)

= 3x3 + 3x2 + x – 9

(c) P(x) + 2Q(x) – 3R(x) = (5x3 – 3x2 + 4x + 7) + 2(6x2 + 5x – 4) – 3(8x3 + 5x – 2)

= (5x3 – 3x2 + 4x + 7) + (12x2 + 10x – 8) – (24x3 + 15x – 6)

= 5x3 – 3x2 + 4x + 7 + 12x2 + 10x – 8 – 24x3 – 15x + 6

= 5x3 – 24x3 – 3x2 + 12x2 + 4x + 10x – 15x + 7 – 8 + 6

= – 19x3 + 9x2 – x + 5

  1. F(x) = 3x3 + 4x2 – 5x + 9
  2. F(-1) = 3(-1)3 + 4(-1)2 – 5(-1) + 9

= 3(-1) + 4(1) – 5(-1) + 9

= -3 + 4 + 5 + 9

= 15

  1. F(1) = 3(1)3 + 4(1)2 – 5(1) + 9

= 3(1) + 4(1) – 5 + 9

= 3 + 4 – 5 + 9

= 11

  1. F(0) = 3(0)3 + 4(0)2 – 5(0) + 9

= 9

  1. F(3) = 3(3)3 + 4(3)2 – 5(3) + 9

= 3(27) + 4(9) – 15 + 9

= 81 + 36 – 15 + 9

= 111

Class activity

  1. Given that P1(x) = 2x2 + 3x + 4; P2(x) = 4x2 – 6x + 8; P3(x) = 5x3 – 3x2 + 5x + 6. Find 4 P1(x) + 5 P2(x) – 2 P3(x).
  2. If f(x) = x4 – 3x3 + x2 + 3x – 2, show that f(1) = f(2).

Multiplication of Polynomials

The multiplication of two polynomials is obtained by using every term of one polynomial to multiply each term of the other polynomial and collecting together like terms.

When a polynomial of degree m is multiplied by another polynomial of degree n, another polynomial of degree m + n is obtained.

Examples

  1. Given P(x) = 7x3 – 4x2 + 3x +4 and Q(x) = 5x2 + 6x +1. Find PQ.

Solution:

Method 1:

PQ = (7x3 – 4x2 + 3x +4)(5x2 + 6x +1)

= 7x3(5x2 + 6x +1) – 4x2(5x2 + 6x +1) + 3x(5x2 + 6x +1) +4(5x2 + 6x +1)

= 35x5 + 42x4 + 7x3 – 20x4 – 24x3 – 4x2 + 15x3 + 18x2 + 3x + 20x2 + 24x + 4

Rearrange

= 35x5 + 42x4 – 20x4 + 7x3 – 24x3 + 15x3 – 4x2 + 18x2 + 20x2 + 24x + 3x + 4

= 35x5 + 22x4 – 2x3 + 34x2 + 27x + 4

Method 2: (Long Multiplication)

7x3 – 4x2 + 3x +4

X 5x2 + 6x + 1

7x3 – 4x2 + 3x +4

42x4 – 24x3 + 18x2 + 24x

35x5 – 20x4 + 15x3 + 20x2

35x5 + 22x4 – 2x3 +34x2 + 27x +4

  1. Given that P(x) = 5x3 + 3x2 – 2x + 4 and Q(x) = 2x2 + 1, find P(x).Q(x)

Solution

Method 1:

PQ = QP = (2x2 + 1) (5x3 + 3x2 – 2x + 4)

= 2x2(5x3 + 3x2 – 2x + 4) + 1 (5x3 + 3x2 – 2x + 4)

= 10x5 + 6x4 – 4x3 + 8x2 + 5x3 + 3x2 – 2x + 4

= 10x5 + 6x4 + x3 +11x2 – 2x + 4

Method 2:

5x3 + 3x2 – 2x + 4

2x2 + 1

5x3 + 3x2 – 2x + 4

10x5 + 6x4 – 4x3 + 8x2

10x5 + 6x4 + 2x3 +11x2 – 2x +4

Class activity

  1. If f(x) = 3x2 – 2x + 3, g(x) = x2 – 3x + 4 and h(x) = x2 + 5, find f(x).g(x).h(x).
  2. Given that P1(x) = 5x3 + 3x2 – 2x + 6; P2(x) = x3 + 4x2 – 3x + 1 and P3(x) = 2x3 – 3x2 + 3x + 2, find (i) (P1 + P2)P3 (ii) (P2 – P3)P1.

Division of polynomials

A polynomial f(x) can be divided by another polynomial g(x), provided the degree of g(x) is not greater than that of f(x).

So, we could divide say x3 + 3x2 – x + 1 (3rd degree), by x-2 (1st degree), but not x – 2 by x3 + 3x2 –x + 1.

If n ≥ m, the result of dividing a polynomial of degree n by a polynomial of degree m is another polynomial of degree n – m.

Suppose we want to divide x3 + 3x2 – x + 1 by x-2, the polynomial x3 + 3x2 – x + 1is called the dividend while the polynomial x-2 is called the divisor. The result of the division is called the quotient and what is left after the division is called the remainder.

Examples

  1. Find the quotient and remainder when x3 + 3x2 –x + 1 is divided by x – 2.
  2. Find the quotient and remainder when 4x3 – x2 + x – 5 is divided by x2 + x – 1.

Solution 1.

Step 1: Divide the first term of the dividend by the first term of the divisor to obtain the first term of the quotient.

X2

x3 + 3x2 – x + 1

x + 1

X – 2

Step 2: Multiply the divisor by the first term of the quotient gotten and write the result under the dividend.

X2

x3 + 3x2 – x + 1

x3 – 2x2

x

X – 2

Step 3: Subtract the product obtained in step 2 from the dividend.

X2

x3 + 3x2 – x + 1

x3 – 2x2

5x2 – x + 1

x

X – 2

Step 4: Repeat steps 1, 2, and 3 with x2 – 5x + 1 as the new dividend.

X2 + 5x

X – 2

x3 + 3x2 – x + 1

x3 – 2x2

5x2 – x + 1

5x2–10x

9x + 1

x

Step 5: Repeat steps 1, 2, and 3 with 9x + 1 as the new dividend.

X2 + 5x + 9

x3 + 3x2 – x + 1

-(x3 – 2x2)

5x2 – x + 1

-(5x2–10x)

9x + 1

-(9x -18)

19

x

X – 2

 

Note: x3 + 3x2 – x + 1 is the dividend.

x – 2 is the divisor.

X2 + 5x + 9 is the quotient.

19 is the remainder.

We can combine the quotient, divisor and remainder to get the polynomial as follows:

x3 + 3x2 –x + 1 = (x – 2) x (x2 + 5x + 9) + 19

Polynomial (P) = Divisor (D) x Quotient (Q) + Remainder(R)

Solution 2

4x – 5

x2 + x – 1 4x3 – x2 + x – 5

-(4x3 + 4x2 – 4x)

  • 5x2 + 5x – 5

-(-5x2 – 5x + 5)

10x – 10

Note that in each of the examples above, the degree of Q = degree of P – degree of D and that the degree of R is one less than the degree of D. The degree of a remainder is one less than that of the divisor.

Class activity

  1. Find the quotients and remainders when P(x) = 6x3 + 4x2 – x + 5 is divided by:
  2. 3x2 + 2x +1
  3. x3 – 3x + 2
  4. A polynomial is divided by x + 1. The quotient is 2x – 3 and the remainder is 3, find the polynomial.

SUB-TOPIC 3

Remainder and factor theorem

The long division method used in the previous sub-topic helps us to determine, not only the quotient but also the remainder.

Consider the example 1 of the previous sub-topic:

Find the remainder when x3 + 3x2 –x + 1 is divided by x – 2.

X2 + 5x + 9

x3 + 3x2 – x + 1

-(x3 – 2x2)

5x2 – x + 1

-(5x2–10x)

9x + 1

-(9x -18)

19

x

X – 2

Hence when P(x) is divided by x – 2, the remainder is 19

Now find P(2).

P(2) = (2)3 + 3(2)2 – (2) + 1 = 8 + 12 – 2 + 1 = 19.

From the above, you would have observed that when p(x) is divided by x – a, the remainder is f(a). This forms the basis of Remainder theorem.

The Remainder theorem states that if a polynomial p(x) is divided by x – a, the remainder is p(a).

More generally, if p(x) is divided by ax + b, then the remainder is p(

A special case of the remainder theorem is when p(x) leaves no remainder when it is divided by x – a. when this happens, we say x – a is a factor of p(x).

This modified theorem is called factor theorem and it is states that if p(a) = 0 then x – a is a factor.

Examples

  1. Find the remainder when 2x2 – 5x + 6 is divided by x – 3.
  2. Determine the values of p and q if (x – 1) and (x + 2) are factor of 2x3 + px2 – x + q.

Solution:

  1. Let f(x) = 2x2 – 5x + 6

Let R be the remainder when f(x) is divided by x – 3

Then,

R = f(3)

F(3) = 2(3)2 – 5(3) + 6

= 2(9) – 15 + 6

= 18 – 15 + 6

= 9

  1. Let f(x) = 2x3 + px2 – x + q

If x – 1 is a factor of f(x), then

F(1) = 0

F(1) = 2(1)3 + p(1)2 – (1) + q

= 2 + p – 1 + q

= p + q + 2 – 1

= p + q + 1

p + q + 1 = 0 …. (1)

If x + 2 is a factor of f(x), then

F(-2) = 0

F(-2) = 2(-2)3 + p(-2)2 – (-2) + q

= 2(-8) + p(4) – (-2) + q

= -16 + 4p + 2 + q

= 4p + q -14

4p + q -14 = 0 … (2)

Solving simultaneously…

Subtract (1) from (2)

3p – 15 = 0

3p = 15

P = 5

Substitute the value of p into (2)

q = 14 – 4p

q = 14 – 4(5)

q = 14 – 20

q = -6

Hence p = 5, q = -6

Class activity

  1. Find the remainders without performing long division when
  2. x3 +5x2 – 3x + 1 is divided by x + 1
  3. 2x3 – 4x2 + x – 3 is divided by x + 2
  4. If (3x – 1) is a factor of the polynomial f(x) = 4x3 – 4x2 – x + p, find the value of the constant p.

SUB-TOPIC 4

Zero polynomials

Given a polynomial function f(x), the value x = a such that f(a) = 0 is called the zero of the polynomial.

A zero of the function f(x) is a root of the equation f(x) = 0.

To obtain the zeros of a polynomial f(x), set f(x) = 0 and solve the equation.

Examples:

  1. Find the zeros of the following polynomials:
  2. F(x) = x2 – 7x + 12
  3. G(x) = x2 – 16
  4. 2 is a factor of 32, find the values of the constants and state the zeros of

Solution 1:

  1. Set f(x) = 0

x2 – 7x + 12 = 0

x2 – 3x – 4x + 12 = 0

x(x – 3) – 4(x – 3) = 0

(x – 3)(x – 4) = 0

x = 3 or x = 4

Hence the zeros of f(x) are 3 and 4.

  1. Set g(x) = 0

x2 – 16 = 0

(x – 4)(x + 4) = 0

x = 4 or x = -4

Hence the zeros of g(x) are 4 and -4.

Solution 2:

If 2 is a factor, it means the zeros of 2 when substituted into will give the complete value of

from 2

2

 

(x + 2) (x – 3) = 0

x = -2 or x = 3

3(3)3 2

3(-2)3 2

⇒

From equation (i),

Substitute

Hence,

3232

To get the third factor of divide by 2

 

2 32

32

  • 2

2

0 0 0

The third factor is ⇒ hence,

More generally, an nth degree polynomial will have n zeros while an equation of degree n will have n roots.

Class activity

Given that a and b are the zeros of the polynomial f(x) = x2 – x – 6 with a˃b, and that g(x) = f(x + 2), find:

  1. g(a) + g(b)
  2. g(a) – g(b)
  3. g(a) x g(b)

PRACTICE QUESTIONS

  1. Given that P(x) = ax2 + bx + 1, P(1) = 6 and P(-1) = 2, determine the values of a and b.
  2. In the identity ax2 + bx + c = x2 – 2, find a, b, c.
  3. Given that f(x) = x3 + 3x2 + 3x + 1 and g(x) = x3 +- 3x2 + 3x -1, find (a) f(x) + g(x) (b) f(x) – g(x) (c) f(x) x g(x)
  4. Given that p(x) = ax2 + bx +1, p(1) = 6 and p(-1) = 2, determine the values of a and b.
  5. Find the quotients and remainders when P(x) = 6x3 + 4x2 – x + 5 is divided by:
  6. 3x2 + 2x +1
  7. x3 – 3x + 2
  8. A polynomial is divided by x + 1. The quotient is 2x – 3 and the remainder is 3, find the polynomial.

EVALUATION

  1. The polynomial has the same remainder when divided by (x+2) and (x-1). Find the value of constant q. (A) -11 (B) -9 (c) -3 (D) -1
  2. The polynomial has a remainder 20 when divided by (x – 2). Find the value of constant P. (A) 8 (B) 6 (C) -6 (D) -8
  3. Find the remainder when is divided by x-2 (A) 28 (B) – 28 (C) – 56 (D) 56
  4. The expression leaves a remainder of 6 when divided by x-k. The positive value of k is (A) 1 (B) 2 (C) 3 (D) 4
  5. Given f(x) = ax3 + 2x2 + bx + c, and f(0) = f(1) = 4 and f(-2) = 8, find the values of a, b and c.
  6. When is divided by (x-1), (x+2) and (x-3) the remainders are -14, 16 and 36 respectively. Find the values of the constants p, q and r and hence determine the quotient and remainder when f(x) is divided by x-4.

WEEK TWO

TOPIC: POLYNOMIALS 2

SUB-TOPICS:

  1. Roots of cubic equation: Sum of roots; Sum of products of two Roots; Products of roots.
  2. Graphs of polynomial function.

SUB-TOPIC 1

The general cubic equation takes the form: 3 (because if it becomes a quadratic equation).

3 …(1)

Dividing through by a,

3 …(2)

Let the roots of this equation 2 be. Then 3

= 0 …(3)

2

2– 2

Collect like terms

…(4)

By comparing coefficients of equations (2) and (4)

  1. The sum of roots
  2. The sum of product of roots
  3. The product of roots

Example 1:

The roots of a cubic equation are such that obtain the equation the roots of which are 2, 2, 2.

Solution:

The three ways of obtaining 2, 2 and 2 include.

(a). Expanding 2

22+2

= 2 22

2+22

⇒ 22+22

2+22 2

2+222

(b). Expanding 2 we have:

2222222

2

(c). 22222

The required equation is 322+2)222+2(22)]2 = 0

3 2

32

Example 2:

One of the roots of the cubic equation.

Find the:

  1. Sum of the two other roots;
  2. Product of the two other roots.

Hence or otherwise, find the other two roots.

Solution

Let α, β and γ be the roots of the equation such that γ = 5, then

Given that

a = 1, b = -9, c = 23, d = 15

hence:

… (1)

Also,

= 15

15

= 3 … (2)

From equation (1) we have

Substituting 4 – for

= 3

If α =1, then β =3 and

If α = 3, then β = 1

Hence:

α=1; β=3,γ=5.

Class activity

  1. The equation 32 has roots Find the equation whose roots are 3,3,3.
  2. Write down the cubic equation with solutions such that and

SUB-TOPIC 2

Graphs of polynomial function

The shape of a polynomial graph depends on the degree of that polynomial.

Polynomials of degree one

The straight line is the graphical representation of polynomials of degree one. The coefficient of x gives us a measure of the gradient or slope of the line.

If a ˃ 0, the straight line rises as y increases when x also increases. If a ˂ 0, the straight line falls as y decreases when x increases. If the graphs below, the points A and B on the straight line are called the x and y intercepts respectively.

y

Y = ax – b

a ˂ 0

B

y

Y = ax +b

a ˃ 0

B

A

x

A

x

Determining x- and y-intercept

To find the x-intercept, put y = 0 and solve for x in the equation. The x-intercept is identified as the zero of the corresponding polynomial.

To find the y-intercept, put x = 0 and solve for y.

From the knowledge of the intercepts, one can easily sketch the graph of a polynomial of degree 1.

Polynomial of degree two

The parabola is the graphical representation of polynomials of degree two. It has two shapes which depends on whether the coefficient of x2 is positive or negative.

Determining x- and y-intercept

To find the x intercept, put y = 0 and solve for x. the values of x for which y = 0 are the zeros of the polynomial.

To find y-intercept, put x = 0.

Turning points

The lowest point A on the curve in graph 1 is a turning point and it is called Minimum point.

The highest point B on the curve in graph 2 is also a turning point and it is called Maximum point.

Polynomials of degree three

The curve of polynomials of degree 3 is usually called cubical parabola and it has two shapes depending on whether a ˃ 0 or a ˂ 0.

Examples:

  1. Sketch y = 2x -1 by first finding the slopes and intercepts on the axes
  2. Sketch y = x2 +2x – 3 showing the intercepts and turning points.
  3. Sketch the curve represented by y = 12 + 4x -3x2 – x3.

Class activity

  1. Show that (2x-1) is a factor of the polynomial f(x) = 8x3 – 8x2 + 1 and find the quadratic factor.
  2. Sketch the graphs of the following: (i) y = -3x + 2. (ii) y = 8 – 2x – x2. (iii) y = x3 + 2x2 – 5x – 6.

PRACTICE QUESTIONS

  1. The expression px2 + qx +6 is divisible by x-3, and has a remainder of 20 when it is divided by x + 1. Find the values of p and q.
  2. When the polynomial f(x) = px3 + qx + r (where p, q and r are constants) is divided by (x + 3) and (x – 2), the remainders are -12 in each case. If (x + 1) is a factor of f(x), find: (i) f(x); (ii) the zeros of f(x).
  3. Given that x – 2 is a factor of 2x3 – x2 – 8x +4, find the other two factors.
  4. The equation 32 has roots Find the equation the roots of which are 3,3,3.
  5. Factorise completely 4x3 – 8x2y – 9xy2 + 18y3.

EVALUATION

  1. If the polynomial x3 + px2 + qx – 6 has a factor (x – 1) and leaves a remainder of -24 when divided by (x + 1):
  2. find the constants p and q.
  3. factorise the polynomial completely and find its zeros.
  4. Factorise 432 and 32 completely.
  5. Write down the cubic equation with solutions such that and
  6. The remainders when f(x) = x3 + ax2 + bx + c is divided by (x – 1), (x + 2) and (x – 2) are respectively 2, -1 and 15, find the quotient and remainder when f(x) is divided by (x + 1).
  7. If the polynomial f(x) = ax2 + 13x = b and g(x) = 4x2 + px + q are divided by x – 1, the remainders are 12 and 16 respectively. It they are divided by x – 2, the remainders are 40 and 20 respectively. Find the values of the constant a, b, p and q and hence determine the values of x for which f(x) = g(x).

WEEK THREE

TOPIC: PERMUTATIONS

SUB-TOPICS:

  1. Permutation (arrangement).
  2. Cyclic permutation.
  3. Arrangement of identical objects.
  4. Arrangement in which repetitions are allowed.

SUB-TOPIC 1

Permutation (Arrangement)

Suppose we are interested in the different arrangement of two people in a line. If these two people are labelled a and b then the problem is the same as finding the different arrangement of the letters a and b.

The number of arrangement will be two i.e. ab and ba.

Suppose there are three people in the line labelled a, b, and c. finding the different arrangements will be the same as finding the arrangements of the letters a, b and c.

c

b

a

c

b

b

a

c

a

b

a

c

c

a

b

bac

bca

abc

cab

cba

acb

From the above, we see there are six different arrangements.

We can get the number of different arrangement of an arbitrary n terms.

NNo. of different arrangementsFormula
111
222 x 1
363 x 2 x 1
4244 x 3 x 2 x 1
51205 x 4 x 3 x 2 x 1

Based on the above pattern, the number of different arrangement of n objects will be:

. This product can be written as n for short. n is read n-factorial.

The factorial of a positive integer is the product of all integers less than or equal to that given number.

I have four balls of different colours: Blue (B), Green (G), Red (R) and Yellow (Y). If I pick three of the balls, the following are the possible results of picking in order:

BGR BRG BGY BYG BRY BYR

GBR GRB GBY GYB GRY GYR

RBG RGB RBY RYB RGY RYG

YBG YGB YBR YRB YGR YRG

TOTAL = 24

Each of these arrangement is called a permutation. For the above, we obtained 24 permutations of four (4) colours taking three at a time. The way this is done is as follows:

The 1st ball could be any of the four balls available;

The 2nd ball could be any of the three colours remaining;

The 3rd ball could be any of the two colours remaining.

Thus, we have. This is called the arrangement of 4 balls taking 3 at a time.

If we have five colours to arrange, taking 5 at a time, we will obtain permutations.

We apply the basic counting principle that: ‘‘If an activity ‘A’ can be performed in m and another activity ‘B’ can be performed in , then, the two activities can be performed one after the other in

Permutation can therefore be defined as each of several possible ways in which a set or number of things can be ordered or arranged. It is also all possible arrangement of a collection of things where the order is important.

Suppose we are only interested in the number of ways, the first and second positions can be taken by 4 people in a race, assuming there is no tie.

The first position can be taken in four ways by any of the four athletes. The second position can be taken in 3 ways by any of the remaining 3 athletes.

So, the number of ways the first and the second positions can be taken by four people in a race is. This arrangement is called permutation of 4 people taking 2 at a time and is denoted by

 

Also, the number of permutations of 8 objects taking 3 at a time is denoted by

 

 

Thus, the general formula is

 

This is the permutation of n objects taking r at a time.

Examples

  1. Evaluate the following: (a) 7! (b) 0! (c) 1!
  2. Simplify
  3. Evaluate each of the following: (a) (b)
  4. Find the number of ways of arranging the letters of the word, EIGHT.

Solution:

  1. (a)

(b) 0! = 1

(c) 1! = 1

2.

  1. (a)

(b)

  1. EIGHT has five different numbers. Hence, the number of permutation is

 

Class activity

  1. Evaluate each of the following: a) b)
  2. In how many ways can five bulbs of different colours be arranged in five socket in a row?
  3. In how many ways can the letters of the word ENGLISH be arranged?

SUB-TOPIC 2

Cyclic permutation

In cyclic permutation, we are concerned about arrangement of this about a circular object.

If the letters A, B, C, D are arranged in that order in a circle, and then A is moved to B’s position and B to C’s position, C to D’s position, and D to A’s position, we obtain the same arrangement. i.e,

A

D

C

B

D

C

B

A

 

To obtain different arrangement, we fix one of the letters and arrange the remaining three in the remaining spaces. This gives 3! arrangements of the 4 letters.

A

D

C

B

A

C

D

B

A

D

B

C

A

C

B

D

A

B

D

C

A

B

C

D

In general, the number of ways of arranging ‘n’ objects in a circle is given by:

No. of ways = 1 x (n-1)!

When beads are threaded in a ‘ring’ the clockwise and the anticlockwise arrangements are not distinguishable and the ring can be turn over.

Thus, the number of distinct arrangements of ‘n’ objects round a circular ring which can be turned over is:

Examples:

  1. In how many ways can 8 boys be arranged at a round table?

Solution:

Total number of arrangements = 1 x (8-1)! = 7! = 5040

  1. Seven beads of different colours are threaded in a ring. How many different arrangement is possible?

Solution:

No. of arrangements =

Class activity

  1. In how many ways can eight boys be arranged around a circular table?
  2. A family of seven is to be seated round a table. In how many ways can this be done if the father and the mother are to sit together?
  3. In how many ways can eight men be seated at a round table if two particular men refused to sit together?

SUB-TOPIC 3

Arrangements of identical objects

Consider the arrangement of 8 bulbs (4 red, 3 blue and 1 yellow) in a row, there are 8! Possible permutations (arrangements). Out of these permutations, 4! Permutations involving changes in position of the red bulbs are not distinguishable and the 3! Permutations of the blue bulbs are also not distinguishable.

Thus, the number distinct permutations of the 8 bulbs (4 red, 3 blue and 1 yellow) =

In general, the number of distinct permutations of the ‘n’ objects containing p of one type of q of a second type and r of a third type is given by:

Example

In how many ways can the letters of the following words be arranged ?

  1. ABAKALIKI
  2. MATHEMATICS

SOLUTION

a) ABAKALIKI 9 letters; letter A appears 3 times, K twice, I twice, B&L once each.

No. of permutation =

b) MATHEMATICS = 11letters ( 2 M’s, 2 T’s, 2 A’s, & others once)

 

Class activity

  1. Find the number of ways the letters of the words FURTHER can be permuted.
  2. In how many ways can the letters of the word STRANGE be arranged so that the vowels occupy only the odd places?
  3. Find the number of arrangement in the letters of the word CONGRATULATIONS if the letter A must be placed next to each other. (Leave your answer in terms of factorial).

SUB-TOPIC 4

Arrangements in which repetitions are allowed

Examples

  1. How many numbers greater than 600 can be formed from the digits 2, 3, 4, 5, 6, 7?
  2. How many four-digits even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 if
  3. No repetitions are allowed
  4. Repetitions are allowed.

SOLUTION

(a) Without repetition: Numbers greater than 600 that can be formed from the option of 6 digits 2, 3, 4, 5, 6, 7 can be a 3-digit number, 4-digit number, 5-digit number or 6-digit number.

For a 3-digit number to be greater than 600, the first digit must be any of the 2 digits (6 or 7)

The 2nd digit, any of the remaining 5 digits ( i.e, after choosing one digit from 6 or 7)

The 3rd digit, any of the remaining 4 digits.

Therefore,

For 3-digits numbers we have

For 4-digits numbers, we have

For 5-digits numbers, we have

For 6-digits numbers, we have

Total = 40 + 360 + 720 + 720 = 1840

With repetition: the numbers of digits greater than 600 that would be formed with 2, 3, 4, 5, 6 and 7 will be:

For 3 digits numbers, there will be

For 4 digits numbers,

For 5 digits numbers,

For 6 digits numbers,

Therefore, total numbers that can be formed = 72+1296+7776+48656 = 57,800

(b) For an even number, the last digit must be 2 or 4 or 6.

Without repetition: The last digit can be any of the 3 digits.

The 1st digit can be any of the remaining 5 digits

The 2nd digit can be any of the remaining 4 digits

The 3rd digit can be any of the remaining 3 digits.

Therefore number required =

With repetition: the number required =

Class activity

  1. How many 4-digits numbers can be formed from the digits 0, 1, 2, 3, …9

If: (a) repetitions are allowed (b) the last digit must not be zero and repetitions are not allowed.

  1. How many 4-digit odd numbers can be formed with the digits 1, 2, 3 and 4 if: (a) repetition is allowed (b) repetition is not allowed.
  2. How many numbers less than 3000 can be formed from the digits 1, 2, 3, 8 and 9, if no digit is used more than once?

PRACTICE QUESTIONS

  1. Simplify. (a) 24 (b) 42 (c) 72 (d) 27
  2. Five students are lined up in a row. How many arrangement could be made if the position of the last boy remains unchanged? (a) 120 (b) 12 (c) 21 (d) 24
  3. Find the number of ways in which the letters of the word STATISTICS could be arranged. (a) 15120 (b) 5120 (c) 2020 (d) 1512
  4. Seven students were late to a class. In how many ways can they occupy: (a) three available vacant seats? (b) Nine available vacant seats?
  5. In how many ways are there of arranging 3 different jobs between 5 men where any man can only do one job?

EVALUATION

  1. In how many ways can 8 people be seated on a bench if only 3 seats are available?
  2. If the mathematics department of a particular college has 5 members of staff and they are to pose for a photograph by standing in a row. How many different arrangements are possible?
  3. Five people are to have a dinner jointly. In how many ways can they sit round a table if a couple must sit together?
  4. How many numbers greater than 4000 can be formed using some or all the digits 6, 5, 4, 3 and 2 without repetition? How many of these will be even?
  5. How many words can be formed from the letters of VALEDICTORY provided the letters A, E, I, O, Y are not to be separated at all?

WEEK FOUR

TOPIC: COMBINATION

SUB-TOPICS:

  1. Combination (selection).
  2. Conditional arrangements and selection.
  3. Probability problems involving arrangement and selection.

SUB-TOPIC 1

Combination (Selection)

In many situations, we make selection without regard to the order. If a committee of 4 members is to be formed from 7 members of staff of DLHS, the order in which the numbers of the given committee are selected is not important.

Combination is therefore a way of selecting items from a collection such that (unlike permutation) the order of selection does not matter.

In selecting three colours from 5 colours: (B, G, R, W, Y), BGR, BRG, GBR, RBG, RGB, are counted as 6 different arrangements (permutations), although they consist of the same 3 colours. The 6 permutations thus represent one combination. Thus, each combination of three objects yields 3! permutation.

Now, the number of the permutations of 5 colours taking 3 at a time, i.e,

The number of combinations of 5 colours taking 3 at a time, i.e,

In general,

Examples

  1. Out of the five science club members of a school, A, B, C, D and E, just three are to be chosen to represent the school in an exhibition. In how many ways can this be done?
  2. In how many ways can a committee of 3 chemistry teachers and 5 mathematics teachers be formed from 6 chemistry teachers and 10 mathematics teachers?

Solution

  1. The three representatives can be selected in
  2. The chemistry teacher can be selected in 6C3 ways and the mathematics teacher can be selected in 10C5 ways.

Total number of ways = 6C3 x 10C5

=

=

=

=

Class activity

  1. In how many ways can a disciplinary committee of 3 be formed from 10 members of staff of a college?
  2. A man travels from Nigeria to Ghana by air and from Ghana to Liberia by ship. He returns by the same means. He has 6 airlines and 4 shipping lines to choose from. In how many ways can he make his journey without using the same airline or shipping line twice?

SUB-TOPIC 2

Conditional arrangement and selection

When restrictions are placed on arrangements or selection, then, the permutation or combination is said to be conditional.

Example 1:

Find the number of ways 6 people can be seated in a round table if two particular friends must sit next to each other.

Solution:

If two people must seat next to each other, the number of ways these friends can sit is 2!

Therefore, the number of ways six people can sit in a round table with two friends that must be together is

Example 2:

A committee of 4 people is to be chosen from 5 married couples. Find how many ways the committee can be chosen if: (i) everyone is equally eligible; (ii) the committee should include at least one woman.

Solution:

i) 5 married couples includes 5 men and 5 women. Since everyone is equally eligible, then, the possible ways of selecting 4 people for the committee are:

4 men and 0 women or

3 men and 1 women or

2 men and 2 women or

1 man and 3 women or

0 men and 4 women.

i.e,

 

 

ii) If at least one woman must be in the committee, then, the possible ways of selecting 4 members of the committee from the couples (5men & 5women) are:

3 men and 1 woman or 2 men and 2 women or 1 man and 3 women or 4 women

i.e,

 

Class activity

  1. An excursion group of 4 is to be drawn from among 5 boys and 6 girls. Find the number of ways of choosing the excursion group if the group:
  2. is to be made up of an equal number of boys and girls;
  3. is to be either all boys or all girls;
  4. has no restrictions on its composition.
  5. A candidate is expected to attempt 12 out of 15 questions. In how many ways can this be done if:
  6. the candidate is to attempt any 12 question;
  7. the first 8 questions are compulsory;
  8. a question is outside the syllabus and hence cannot be completed.

SUB-TOPICS 3

Probability problems involving arrangement and selection

Example 1:

A box contains 10 red, 3 blue and 7 black balls. If three balls are drawn at random, what is the probability that: (a) all 3 are red, (b) all 3 are blue, (c) one of each colour is drawn?

Solution:

Number of ways of selecting any 3 balls from 20 balls = number of element in the sample space.

Number of ways of selecting 3 red balls out of 10 =

(a) P(all the 3 balls are red) =

(b) P(all 3 are blue) =

(c) P(1 red, 1 blue & 1 black) =

Example 2:

Three-digit numbers are formed from the digits 1, 2, 4, 5 and 6. If repetition is not allowed and a number is picked at random, find the probability that it is a multiple of 5 or an odd number.

Solution:

Since probability is involved, we find the sample space for 3-digit numbers formed from 5-digits i.e 5P3.

n(sample space) =

  1. To get multiple of 5, the last digit must be 5. That implies that

The 1st digit can be any of the remaining 4 digits.

The 2nd digit can be any of the remaining 3 digits

No. of ways = 1 x 4 x 3 = 12 ways.

Pr(multiple of 5) =

  1. To get odd number, the last digit will be any of 1 and 5

The first digit can be any of the remaining four

The second digit will be any of the remaining three.

No. of ways = 2 x 4 x 3 = 24ways.

Pr(odd number) =

Hence, pr(multiple of 5 or odd number) =

=

=

Class activity

  1. How many committee of size 5 consisting of three men and 2 women can be selected from 8 men and 6 women if a certain man must not be in the committee?
  2. A bag contains 5 white, 2 black and 3 green balls. If three ball are drawn at random, find the probability that:
  1. All three are green
  2. All three are white
  3. 2 are white and 1 is black
  4. At least, one is black
  5. 1 of each colour is drawn.

PRACTICE QUESTION

  1. How many committee of size 5 consisting of 3 men and 2 women can be selected from 8 men and 6 women if a certain man must not be in the committee? (a) 315 (b) 525 (c) 840 (d) 1287
  2. In how many ways can 9 bulbs be selected from 4 red, 5 green and 6 yellow bulbs if 3 of each colour are to be selected? (a) 800 (b) 120 (c) 40 (d) 27
  3. The number of ways of arranging 9 men and 8 women in a row, when the women occupy the even places is — (a) (b) (c)!8! (d)
  4. A panel consists of 5 men and 4 women. What is the probability of 4 men and 2 women?

(a) (b) (c) (d) .

5. Five digit numbers are formed from digits 4, 5, 6, 7 & 8

  1. How many of such numbers can be formed if repetition of digit is (i) allowed (ii) not allowed?
  2. How many of the numbers are odd if repetition of digits is not allowed?

EVALUATION

  1. Find the number of ways of arranging 9 men and 8 women in a row, if the women occupy the even places.
  2. If find the value of n.
  3. A panel of 5 jurist is to be chosen from a group of 6men and 7 women. Find the number of different panels that could be formed if: (a) a particular man must serve on the panel (b) there is no restriction.
  4. A business man intends to give a dinner party for 6 of his 10 friends. If 2 of them will not attend the party together, in hoe many ways can he select his guests?
  5. A family of 7 is to be seated round a table. In how many ways can this be done, if the father and the mother are to sit together?
  6. Four delegates are to be chosen from 8 members of staff of a college. If 2 of them are senior members of staff, how many different delegations are possible if: (i) only one of the senior members of staff must be in the delegation? (b) the two senior member of staff must be included?

WEEK FIVE

TOPIC: BINOMIAL EXPANSION 1

SUB-TOPICS:

  1. Pascal triangle.
  2. Binomial expansion of (a+b)n , where n can be positive integer, negative integer or fractional value.

SUB-TOPIC ONE

The Pascal’s triangle is a format for getting the coefficients of expansions. It applies to binomial and binomial formed from a reduced polynomial.

Consider the expansion of each of the following:

(a+b)0

(a+b)1

(a+b)2

(a+b)3

(a+b)4

(a+b)5

By ordinary expansion of algebraic terms, we have:

(a+b)0 = 1

(a+b)1 = 1a + 1b

(a+b)2 = 1a2+2ab+1b2

(a+b)3 = 1a3+3a2b+3ab2+1b3

(a+b)4 = 1a4+4a3b+6a2b2+4ab3+1b4

(a+b)5 = 1a5+5a4b+10a3b2+10a2b3+5ab4+1b5

Consider the array of coefficients of a and b. We can display it as follows:

 

n (power)

 

1 0

1 1 1

Row 1 2 1 2

1 3 3 1 3

1 4 6 4 1 4

1 5 10 10 5 1 5

We call the array of coefficients displayed above Pascal triangle named after the celebrated French Mathematician and Physicists Blaise Pascal (1623-1662) noted for his essay on conic section in 1640 and first invention of calculating machine in 1642.

Note the following:

  • A general binomial is of the form (a+b)n .
  • There are n + 1 terms.
  • The expansion is homogenous i.e.the sum of the powers of a and b in each term of the expansion is n.
  • As the power of a descends (starting from n till it reaches 0), the power of b ascends (starting from 0 till it reaches n) and vice versa.

Examples:

  1. Using Pascal’s triangle, expand

Solution:

Let n=4

1 4 6 4 1

Understand that the index (power) of descends as that of ascends.

 

  1. Use Pascal’s triangle to obtain the value of (1.025)4, correct to three decimal places.

Class activity

  1. .

SUB-TOPIC 2

Binomial Expansion of (a+b)n

Binomial Theorem

ncr

ncr =

ncr = ncr

The Binomial Theorem for a positive Integral Index

If a and b are any numbers and n is a positive integer, then

nC0an+nC1an-1b+nC2an-2b2+…+nCran-rbr+…nCnbn

Note:

  1. The number of terms in the expansion is n+1. That is one more than the index of binomial.
  2. The (r+1) term in the expansion of the binomial is called the general term and denoted by Tr+1 =nCran-rbr

The Binomial Theorem for Negative and Fraction

When n is not a positive integer, the expansion becomes

 

Provided a is numerically less than unit. That is -1<< 1. This means that the various coefficients cannot be expressed as nC0, nC1, nC2 etc because they have no meaning when n is not a positive integer.

Again, the theorem can be applied only when the first term of the binomial is unity. If not, the binomial must first be reduced to this form. For example, to expand (t+a)n has to be put in this form .

Examples:

4C0 + 4C1 + 4C2 + 4C3 + 4C4

=

  1. Expand to five terms.

Solution:

Recall:

n = -3,

  1. Expand to four terms.

Solution:

Note the first term of the binomial is not unity. Let’s reduce to the form

n= ½,

Class activity

  1. Determine the coefficient of from the expansion of .
  2. Expand to five terms.
  3. Expand to three terms.

PRACTICE QUESTIONS

  1. Using the binomial theorem, expand (1 + 2x)5, simplifying all the terms. Hence calculate the value of (1.02)5 correct to six significant figures.
  2. If the first three terms of the expansion of the expansion of (1+px)5 in ascending powers of x are 1+20x+160x2 find the values of n and p.
  3. (a) Write down the binomial expansion of (1+y)8, simplifying all the terms.

(b) Using the substitution y = x – x2 in (a), deduce the expansion of (1+ x – x2)8 in ascending powers of x as far as the term in x4.

(c) Find, by inspection, a value of x such that 1+ x – x2 = 1.09. Hence, evaluate (1.09)8 correct to three decimal places.

  1. Write down the binomial expansion of , simplifying all its coefficients.
  2. Obtain the first five terms of the expansion of

ASSIGNMENT

        1. The _______ triangle is a format for getting the coefficient of expansions. (a) Binomial (b) right-angled (c) Pascal’s (d) array.
        2. Pascal’s triangle was named after the French Mathematician and Physicists called_________ (a) Newton (b) Blaise Pascal (c) Cramer (d) Laplace
        3. The general binomial is of the form ______ (a) (a+b)n (b) (a+b)r (c) (a+b)! (d) aCb
        4. The (r+1) term in the expansion of the binomial is called the ________ term. (a) Simple (b) Difficult (c) Binomial (d) General.
        5. Expand
        6. Expand.

WEEK SIX

TOPIC: BINOMIAL EXPANSION 2

SUB-TOPICS:

  1. Finding nth term.
  2. Application of binomial expansion

SUB-TOPIC 1

Finding the nth term

The binomial expansion of is given as:

=

This expansion is true for any natural number value of n, large or small, but when n is a large natural number then is small. Indeed, the larger the value of n, the closer becomes to zero. The n notation for this,

 

the limit of as n→ is 0.

Also, as n→, the closer the expansion above becomes to the sum of terms:

Here the ellipsis (…) at the end of the expansion means that the expansion never ends, that is it has infinite number of terms.

Now we can use the sigma notation and write:

 

Notice the symbol for infinity (∞) at the top of the sigma, this denotes the fact that the sum is a sum of an infinite number of terms.

Examples:

  1. How many terms will there be in the expansion of. Find the fourth term.

Solution:

n=11, x=2x and

  1. Number of terms = n+1 = 11+1 =12
  2. From the formula,

11-3r = 2

-3r =2-11

-3r = -9; 3r = 9 ; hence r=3 (i.e. the 4th term)

 

  1. Find the term x2 and then term independent of x in the expansion of find the fifth term:

Solution:

  1. The general term is

When it is x2, then 12-2r = 2

r=5 (i.e. the sixth term)

Hence,

=

=

=

  1. If the term is to be independent of x, then 12-2r = 0

r =6 (i.e. the 7th term)

= = =

The independent term of x is

Class activity

  1. Find the term independent of x in the expansion
  2. How many terms will there be in the expansion?

SUB-TOPIC 2

Application of Binomial expansion in approximation

If x is very small, we can take 1 + nx as an approximation of is taken as 1-nx.

Examples:

Find the linear approximation of the following:

  1. (1.01)5 (b) (1.02)4 (c) (1.05)3 (d) (0.98)5

Solution: (a) (1.01)5

Expansion of (1+x)n = 1+nx= (1+0.01)5=1+5×0.01= 1+0.050= 1.050

  1. (1.02)4

Solution: Expansion of (1+x)n = 1+nx

(1.02)4 = (1+0.02)4≈1+4×0.02

= 1+0.080= 1.080

(c) (1.05)3; Expansion of (1+x)n = 1+nx

(1.05)3 = (1+0.05)4≈1+3×0.05= 1+0.15= 1.150

(d) (0.98)5

Solution: Expansion of 1+xn = 1+nx

(0.98)5 = (1-0.02)5≈1-5×0.02= 1-0.1= 0.09

EVALUATION

Find the linear approximation of the following:

  1. (1.002)15 (ii) (0.99)8 (iii) (2.004)7

PRACTICE QUESTIONS

  1. How many terms will there be in the expansion of
  2. Find the terms in the expansion of . Find the fift term.
  3. Expand and hence evaluate (0.98)7 correct to three decimal places.
  4. Find the linear approximation of the following (a) (2.004)7 (b) (1.003)11 (c) (d) .
  5. Using the binomial theorem, write down and simplify the first seven terms of the expansion of (1+2x)10 in ascending powers of x. use your expansion to show that 1.210˃6.19.

WEEKEND ACTIVITY:

  1. Find the number of terms in the expansion of . Find the third term.
  2. Find the term x2 and independent of x in the expansion .
  3. Find the coefficient of x-4 in the expansion.
  4. Determine the coefficient of from the expansion of .
  5. Expand to four terms.

WEEK SEVEN

MID TERM BREAK.

WEEK EIGHT

TOPIC: ROOTS OF QUADRATIC EQUATION 1

SUB-TOPICS:

  1. Quadratic equation (completing the square and formula method).
  2. Sum and product of roots of quadratic equation.
  3. Finding quadratic equation given sum and product of roots, x2 – (sum of roots) + (product) = 0.
  4. Condition for quadratic equation to have: (i) Equal roots (b2 = 4ac) (ii) Real roots b2 > 4ac (iii) No roots b2 < 4ac

SUB-TOPIC 1

Quadratic equation (completing the square and formula method).

A quadratic equation (trinomial) in one variable is a three termed equation in which the highest power of the variable is two. The general quadratic equation in variable x is of the form ax2 + bx + c = 0 where a ≠ 0.

In general, a quadratic equation has two solutions which may or may not b+e equal.

There are four major methods of solving quadratic equations. They are:

  • factorisation method;
  • completing the square method;
  • formula method;
  • graphical method.

Factorisation

Examples

Solve the following by factorisation

Solution

(a) Product of 1st and 3rd term = -8x2

Factor of -8x that will sum to -2x (the middle term) = -4x and +2x

Replacing the middle term with the two factors, we have:

(b) Product = -120x2

Sum = 14x

Factors = -6x and +20x

Completing the square method

Given ax2 + bx + c = 0, a ≠ 0

Write the equation as ax2 + bx = -c

Divide each term by the coefficient of x2. Find half of the new coefficient of x, square it and add this to both sides of the equality sign. This makes the expression on the LHS a perfect square.

Take the square root of both sides and determine the values of x to the desired accuracy.

The above is the process of solving quadratic equation using the method of completing the square.

Quadratic formula

The formula derived above is the quadratic formula.

Example:

Use the quadratic formula to solve

Solution:

a = 2; b = 7; c = -15

Class activity

  1. Use the three methods discussed above to solve the following quadratic equations:
  2. 2x2 – 17x – 9 = 0
  3. 3x2 + 10x – 12 = 0
  4. 8x2 + 34x + 21 = 0

SUB-TOPIC 2

Sum and product of roots of quadratic equation

Recall the completing the square method and general formula ‘ of the general equation

In this case the two roots of the equation are = or

It is possible to represent the two roots by then we have

 

 

 

 

 

 

Product of roots

 

From the above, if we have

We can rewrite this as .

Also, if α and β are the roots of the quadratic equation, then;

 

 

 

Therefore,

and

Given a quadratic equation, we can find the sum and product of the roots using the above information.

Example 1: if the roots of the equation are

Find

Solution: In the equation

 

The roots of the equation are

Then (a)

(b)

(c)

From (a)

 

(d)

 

 

24

= 40

Example 2:- if the roots of the equations

are (c) α2 – β2

Solution :- (i) The equation is

 

 

 

 

has no expression, then

 

 

 

But we want

 

 

 

 

 

This cannot be determined

The equation is

 

 

 

Following example I (c) we can express this as

(

 

 

 

Example 3:- Express the following in terms of

 

Solution:-

  1. .

 

 

.

 

 

Class activity

  1. are the roots of the equation

Find

 

 

  1. Given , prove that

 

 

SUB-TOPIC 3

Find quadratic equation given sum and product of roots

We have been able to establish that given α and β as the roots of a quadratic equation where a, b and c are constant and, then,

 

Also, and

Therefore, given the roots of a quadratic equation, the equation can be gotten this way:

Example 1:-

Given that the roots of equation are 3 and 7, find the equation.

Solution:

Roots of the equation are 3 and 7

Sum of roots

Product of roots (

 

 

Example 2:- Find the equation whose roots are -8 and 2.

Solution: Roots of the equation are -8, 2

 

 

The equation is

 

Example 3: If the roots of the equation are, find the equation.

Solution: Roots are

 

 

 

 

Multiply through by 12

 

Class activity

  1. Construct and simplify equations whose roots are given below:

 

  1. , write out the equation whose roots are . Find the values of from your equation.

SUB-TOPIC 4

Condition for quadratic equation to have:

 

In this section, we want to see some properties of roots. We can determine the type of roots that a particular equation will have.

The quadratic equation of

The part of the roots under square root sign is called the discriminant (i.e.) of the roots of the quadratic equation. This is because it can be used to determine the nature of the roots.

 

This means that whatever is contained inside the square root.

 

The quadratic is said to have coincident roots. This happens when the quadratic equation is a perfect square.

Example 1:- Consider the equation

SOLUTION: By factorization

 

 

 

 

(ii) Real roots: When b2 is greater than 4ac then the value under the roots sign is a positive number that is . The square root of such number will have two real values, one positive and other negative. The equation is said to have two distinct real roots.

Example 2: Solve the equation

 

,

 

 

 

(iii) Imaginary roots: When b2 is less than 4ac then the value under the root sign is a negative number i.e This shows that the root is not a real number. We say that the equation has imaginary roots.

Example 3: Find the roots of the equation

Solution: a=1 , b=2 c=5

Then

 

The roots are imaginary roots

Example 4:- Without solving the equation determine whether the equation has two different roots, coincident roots or imaginary roots.

Solution:- Rewrite the equation

 

Then a=5, b=-5, c=1

Then b2 = 25

 

 

Since

Then the equation has two real different roots.

Example 5:- Determine the nature of the roots equations without solving them.

 

Solution :

Rewrite the equation

 

To find the nature of the roots

 

 

 

 

The equation has two distinct real roots.

 

 

 

 

The equation has no real roots.

Class activity

Determine the nature of the roots of the following equations without necessarily solving them:

 

  1. Two real and different roots?
  2. Coincident and real roots
  3. No roots

PRACTICE QUESTIONS

        1. Given that are the roots of an equation such that the equation..
        2. If the equation has coincidental roots, find the value of P. (a)
        3. If
        4. If ( where p and q are constant, find the possible values of q.
        5. The roots of the equation are α and β, with α greater than β. Find the values of: (i) α – β (ii) α2 – β2.
        6. If the roots of the equation 4 Find the equation whose roots are

ASSIGNMENT

The roots of the equation are where find the value of k.

Find without necessarily solving the equation, the nature of the roots of the equation 3 . The equation (a) distinct two roots. (b) has no roots (c) coincident real roots (d) none of the above

If are the roots of the equation find (a)

Form the equation whose roots are -4 and 9.

The roots of the equation are, where m is a constant. If find the value of m.

If the sum of the squares of the roots of the equation is 1, show that

WEEK NINE

TOPIC: ROOTS OF QUADRATIC EQUATIONS 2

SUB-TOPICS:

  1. Quadratic functions (Simultaneous Equations: One linear, One quadratic)
  2. Solution of problems on roots of quadratic equation.
  3. Maximum and minimum values.

SUB-TOPIC 1

Quadratic functions (Simultaneous Equations: One linear, one quadratic)

We have discussed different ways but we need to mention that graphical solution is very important aspect of solving quadratic equations. This is because with graphical solution a lot of other problems can be solved.

The graph of the quadratic equation called parabola. Some call it cup or cap. The quadratic expression is equated to y and it is called a quadratic function. The example below show the graphical solution of quadratic function.

Example 1: Solve graphically, the equation 2

Solution:

Draw the table of values for the equation 2

01234
22712303122748
3210-1-2-3-4
-2-2-2-2-2-2-2-2
28124-2082242

Choose a convient scale, on and, on let represents and on represents

From the graph we find the point here the curve intersects

The graph is also useful to determine the minimum value of the minimum value of we have minimum point when and maximum point when

Simultaneous Equations

When solving simultaneous equation (you are already used to solving it graphically). In situation where one equation is linear and the second is quadratic, it can be solved by substitution as well as solving graphically.

In graphical solution of one linear-one quadratic simultaneous equation, there are three possible relationships between the straight line (linear) and the parabola (quadratic). They are:

  • Line intersecting with curve
  • Line touching curve at a point (tangent)
  • Line not intersecting the curve.

Example 2: Solve the simultaneous equations: 22

Solution: By substitution:

22

22

Since, then

Hence, adding (iii) together we get

From (ii) .

Example 3: Given the simultaneous equations:

2 and

Show on the graph the points of interest. Hence write out the values of .

Solution:

2

 

Table of values for 2 and

2

2362516941014

From the above, there was no intersect of the curve and the straight line. The solutions to the two equations cannot be determined because there is no point of intersection.

The points of intersection give the solution.

Example 3: On the same axes, plot the graph of y = 2x2-5 + 4 and y = 2x + 3. Hence find the points of intersection of the two graphs.

Solution:

Prepare the table of values for the functions given above.

y = 2x2-5 + 4

X-4-3-2-10123456
2x232188202818325072
-5x20151050-5-10-15-2025-30
+4+4+4+4+4+4+4+4+4+4+4+4
Y563722114127162946

Choose a convenient scale.

The points of intersections x = 0.2 and 3.3

The above example shows the case of the line intersecting with the curve.

Example 4: solve the simultaneous equation y = x2-2x + 2 and y = 4x -7. Interpret your result geometrically.

Solution:

Eliminate y to obtain: x2-2x + 2 = 4x -7 ⇒ x2 – 6x + 9 = 0

By factorisation:

(x – 3)(x – 3) = 0⇒x = 3(twice).

From y = 4x – 7 = 4(3) -7 = 5. The solution is x = 3 and y = 5.

Draw the graphs of the two equation to interpret it geometrically.

Table of values for y = x2 – 2x + 2. y = 4x – 7

x-2-101234x0234
x2410149164x081216
-2x820-2-4-6-8-7-7-7-7-7
+2+2+2+2+2+2+2+2y-7159
y145212510

 

The line y = 4x -7 intersects the curve y = x2 + 2x + 2 at only one point. Therefore, the solution to the equations is at the point x = 3 and y = 7.

Class activity

  1. Solve the simultaneous equations y = 4x – 1 and y =2x2 graphically and interpret your result geometrically.
  2. Solve for -1. Using a scale of 2cm to 1 unit on the x-axis and 2cm to represent 5 units on the y-axis.

SUB-TOPIC 2

Solutions of problems on roots of quadratic equation

Mathematics is important of life situation because of its application. You are used to problems leading to simple equations. We want to see the word problems leading to quadratic equations.

In order to solve such problems, you must take note of the following:

  1. Express the ideas involved in mathematical symbols.
  2. Write out the equation using the symbols.
  3. Solve the equation.
  4. Interprete your result.

Example 1: the product of two consecutive whole numbers is 506. Find the numbers.

Solution:

Let the numbers be x and (x + 1).

Then, x(x+1) = 506 ⇒ x2 + 1 = 506 (this is now quadratic equation)

X2 + x – 506 = 0

Solve by formula to find the values of x using the parameters below

a = 1 b = 1 c = -506

Example 2: There are two possible routes from Lagos to Ijebu Ode. One route is through Lagos/Ibadan express way which is 100km and the other is through Ikorodu-Epe covering a distance of 80km. A motorist going through express way can travel 10km per hour faster than the one going through Ikorodu and Epe and arrive Ijebu-Ode 5 minutes earlier as well. What is the time spent on the journey to Ijebu Ode by the motorist travelling through the express way?

Solution:

Let x be the speed of motorist going through Ikorodu/Epe and the speed of the one going through express way is x + 10.

Time taken by Ikorodu/Epe = 80/x.

Time taken by express way = 100/ (x + 10)

Hence, 80/x – 100/(x+10) = 1/12.

 

Form a quadratic equation from (i) above and solve it using formula and conclude.

Class activity

  1. The length of a rectangular field is 6m more than the width. If the area of the field is 72m2, find the dimensions of the field.
  2. Two consecutive odd integers are such that the sum of their reciprocals is Find the odd integers.

SUB-TOPIC 3

Maximum and Minimum values

The graph of as we have seen is a parabola. We have minimum point when and maximum point when

The maximum or minimum value (y) is

The curve is symmetrical about the line which is called the axis of symmetry.

If f(x) = 0, then,

      1. the curve cuts the horizontal axis if
      2. the curve touches the horizontal axis if
      3. the curve does not cut the horizontal axis if

Example 1:

Find the minimum value of and the corresponding the value of x for which y is a minimum.

Solution:

 

 

 

When x = -5/6, the expression in the brackets will be zero, hence the minimum is -49/12.

The corresponding value of x for which y is minimum is -5/6.

Note that x = -5/6 is the axis of symmetry of the parabola. Alternative, let the minimum value of y be ym then

 

Also the equation of the line of symmetry is

X = -b/2a = -5/6.

General evaluation:

  1. Solve the equations simultaneously and show the points of intersections

Y = 4 – 11x and y = 2x2-19

  1. Find the maximum value of y = 5 + 4x – x2 and the coordinates at the point where the curve y = 5 + 4x – x2, cuts the coordinates axes.
  2. The formula gives the sum of consecutive whole numbers. If
  3. A father got his first son at 31 years. If the product of their ages is 816. Find the ages of the father and his son.

SUBJECT: FURTHER MATHEMATICS CLASS: SSS2

SCHEME OF WORK

WEEK TOPICS

  1. Polynomials 1: (a) Definition of polynomials (b) Basic operations on polynomials. (c) Remainder and factor theorem. (d) Zeros of polynomials.
  2. Polynomials 2: (a) Roots of cubic equation: Sum of roots; Sum of products of two Roots; Products of roots. (b) Graphs of polynomial function.
  3. Permutations: (a) Permutation (arrangement) (b) Cyclic permutation (c) Arrangement of identical objects. (c) Arrangement in which repetitions are allowed.
  4. Combination: (a) Combination (selection). (b) Conditional arrangements and selection. (c) Probability problems involving arrangement and selection.
  5. Binomial Expansion 1: (a) Pascal triangle (ii) Binomial expansion of (a+b)n , where n can be positive integer, negative integer or fractional value.
  6. Binomial Expansion 2: (a) Finding nth term (b) Application of binomial expansion
  7. Mid-term break.
  8. Roots of quadratic equation 1: (a) Quadratic equation (completing the square and formula method). (b) Sum and product of roots of quadratic equation (c) Finding quadratic equation given sum and product of roots, x2 – (sum of roots) + (product) = 0 (d) Condition for quadratic equation to have: (i) Equal roots (b2 = 4ac) (ii) Real roots b2 > 4ac (iii) No roots b2 < 4ac
  9. Roots of quadratic equation 2: (a) Conditions for given line to: (i) intersect a curve, (ii) be tangent to curve, (iii) not intersect a curve (b) Solution of problems on roots of quadratic equation. (c) Maximum and minimum values.
  10. Revision.
  11. Examination.

WEEK ONE

TOPIC: POLYNOMIALS 1

SUB-TOPICS:

(a) Definition of polynomials.

(b) Basic operations on polynomials.

(c) Remainder and factor theorem.

(d) Zeros of polynomials.

SUB-TOPIC 1

Definition of Polynomials

A polynomial is a mathematical expression which is a sum of terms, each term consisting a variable or variables raised to a power and multiplied by a coefficient. A polynomial of one variable x (univariate) has the following as its general form:

anxn + an-1xn-1 + … + a2x2 + a1x + a0

where the highest power of the variable n is the degree of the polynomial; the numerical constants an, an-1, … a2, a1 are called the coefficients of the polynomial, while a0 is called the constant term.

Examples of polynomials include:

  • 3x2 – 2x + 4
  • 2x3 + 3x2 + 5x + 3

A function whose values are given by a polynomial is called a polynomial function. Eg: f(x) = 2x3 + 3x2 + 5x + 3

An equation that is obtained when we set a polynomial equal to zero is called a polynomial equation. E.g.: 2x3 + 3x2 + 5x + 3 = 0

Equality of polynomials

Two polynomials,

P(x) = anxn + an-1xn-1 + … + a2x2 + a1x + a0 and

Q(x) = bnxn + bn-1xn-1 + … + b2x2 + b1x + b0

are said to be equal if:

an = bn

an-1 = bn-1

a2 = b2

a1 = b1

a0 = b0

The value that is obtained by substituting a for x in a polynomial P(x) is denoted by P(a).

SUB-TOPIC 2

Basic operations on polynomial

Addition and Subtraction of Polynomials

Given

P(x) = anxn + an-1xn-1 + … + a2x2 + a1x + a0 and

Q(x) = bnxn + bn-1xn-1 + … + b2x2 + b1x + b0

Then,

P(x) + Q(x) = (an + bn) xn + (an-1 + bn-1) xn-1 + … + (a2 + b2) x2 + (a1 + b1) x + a0 + b0

Similarly,

P(x) + Q(x) = (an – bn) xn + (an-1 – bn-1) xn-1 + … + (a2 – b2) x2 + (a1 – b1) x + a0 – b0

Examples:

  1. Given P(x) = 5x3 – 3x2 + 4x + 7; Q(x) = 6x2 + 5x – 4; R(x) = 8x3 + 5x – 2.

Find, (a) P(x) + Q(x); (b) R(x) – P(x); (c) P(x) + 2Q(x) – 3R(x).

  1. If F(x) = 3x3 + 4x2 – 5x + 9, find: (a) F(-1). (b) F(1). (c) F(0). (d) F(3).

Solution:

  1. (a) P(x) + Q(x) = (5x3 – 3x2 + 4x + 7) + (6x2 + 5x – 4)

= 5x3 + (– 3x2 + 6x2) + (4x + 5x) + (7 + (-4))

= 5x3 + 3x2 + 9x + 3

(b) R(x) – P(x) = (8x3 + 5x – 2) – (5x3 – 3x2 + 4x + 7)

= (8x3 – 5x3) + (0 – (-3x2)) + (5x – 4x) + ((-2) -7)

= 3x3 + 3x2 + x – 9

(c) P(x) + 2Q(x) – 3R(x) = (5x3 – 3x2 + 4x + 7) + 2(6x2 + 5x – 4) – 3(8x3 + 5x – 2)

= (5x3 – 3x2 + 4x + 7) + (12x2 + 10x – 8) – (24x3 + 15x – 6)

= 5x3 – 3x2 + 4x + 7 + 12x2 + 10x – 8 – 24x3 – 15x + 6

= 5x3 – 24x3 – 3x2 + 12x2 + 4x + 10x – 15x + 7 – 8 + 6

= – 19x3 + 9x2 – x + 5

  1. F(x) = 3x3 + 4x2 – 5x + 9
  2. F(-1) = 3(-1)3 + 4(-1)2 – 5(-1) + 9

= 3(-1) + 4(1) – 5(-1) + 9

= -3 + 4 + 5 + 9

= 15

  1. F(1) = 3(1)3 + 4(1)2 – 5(1) + 9

= 3(1) + 4(1) – 5 + 9

= 3 + 4 – 5 + 9

= 11

  1. F(0) = 3(0)3 + 4(0)2 – 5(0) + 9

= 9

  1. F(3) = 3(3)3 + 4(3)2 – 5(3) + 9

= 3(27) + 4(9) – 15 + 9

= 81 + 36 – 15 + 9

= 111

Class activity

  1. Given that P1(x) = 2x2 + 3x + 4; P2(x) = 4x2 – 6x + 8; P3(x) = 5x3 – 3x2 + 5x + 6. Find 4 P1(x) + 5 P2(x) – 2 P3(x).
  2. If f(x) = x4 – 3x3 + x2 + 3x – 2, show that f(1) = f(2).

Multiplication of Polynomials

The multiplication of two polynomials is obtained by using every term of one polynomial to multiply each term of the other polynomial and collecting together like terms.

When a polynomial of degree m is multiplied by another polynomial of degree n, another polynomial of degree m + n is obtained.

Examples

  1. Given P(x) = 7x3 – 4x2 + 3x +4 and Q(x) = 5x2 + 6x +1. Find PQ.

Solution:

Method 1:

PQ = (7x3 – 4x2 + 3x +4)(5x2 + 6x +1)

= 7x3(5x2 + 6x +1) – 4x2(5x2 + 6x +1) + 3x(5x2 + 6x +1) +4(5x2 + 6x +1)

= 35x5 + 42x4 + 7x3 – 20x4 – 24x3 – 4x2 + 15x3 + 18x2 + 3x + 20x2 + 24x + 4

Rearrange

= 35x5 + 42x4 – 20x4 + 7x3 – 24x3 + 15x3 – 4x2 + 18x2 + 20x2 + 24x + 3x + 4

= 35x5 + 22x4 – 2x3 + 34x2 + 27x + 4

Method 2: (Long Multiplication)

7x3 – 4x2 + 3x +4

X 5x2 + 6x + 1

7x3 – 4x2 + 3x +4

42x4 – 24x3 + 18x2 + 24x

35x5 – 20x4 + 15x3 + 20x2

35x5 + 22x4 – 2x3 +34x2 + 27x +4

  1. Given that P(x) = 5x3 + 3x2 – 2x + 4 and Q(x) = 2x2 + 1, find P(x).Q(x)

Solution

Method 1:

PQ = QP = (2x2 + 1) (5x3 + 3x2 – 2x + 4)

= 2x2(5x3 + 3x2 – 2x + 4) + 1 (5x3 + 3x2 – 2x + 4)

= 10x5 + 6x4 – 4x3 + 8x2 + 5x3 + 3x2 – 2x + 4

= 10x5 + 6x4 + x3 +11x2 – 2x + 4

Method 2:

5x3 + 3x2 – 2x + 4

2x2 + 1

5x3 + 3x2 – 2x + 4

10x5 + 6x4 – 4x3 + 8x2

10x5 + 6x4 + 2x3 +11x2 – 2x +4

Class activity

  1. If f(x) = 3x2 – 2x + 3, g(x) = x2 – 3x + 4 and h(x) = x2 + 5, find f(x).g(x).h(x).
  2. Given that P1(x) = 5x3 + 3x2 – 2x + 6; P2(x) = x3 + 4x2 – 3x + 1 and P3(x) = 2x3 – 3x2 + 3x + 2, find (i) (P1 + P2)P3 (ii) (P2 – P3)P1.

Division of polynomials

A polynomial f(x) can be divided by another polynomial g(x), provided the degree of g(x) is not greater than that of f(x).

So, we could divide say x3 + 3x2 – x + 1 (3rd degree), by x-2 (1st degree), but not x – 2 by x3 + 3x2 –x + 1.

If n ≥ m, the result of dividing a polynomial of degree n by a polynomial of degree m is another polynomial of degree n – m.

Suppose we want to divide x3 + 3x2 – x + 1 by x-2, the polynomial x3 + 3x2 – x + 1is called the dividend while the polynomial x-2 is called the divisor. The result of the division is called the quotient and what is left after the division is called the remainder.

Examples

  1. Find the quotient and remainder when x3 + 3x2 –x + 1 is divided by x – 2.
  2. Find the quotient and remainder when 4x3 – x2 + x – 5 is divided by x2 + x – 1.

Solution 1.

Step 1: Divide the first term of the dividend by the first term of the divisor to obtain the first term of the quotient.

X2

x3 + 3x2 – x + 1

x + 1

X – 2

Step 2: Multiply the divisor by the first term of the quotient gotten and write the result under the dividend.

X2

x3 + 3x2 – x + 1

x3 – 2x2

x

X – 2

Step 3: Subtract the product obtained in step 2 from the dividend.

X2

x3 + 3x2 – x + 1

x3 – 2x2

5x2 – x + 1

x

X – 2

Step 4: Repeat steps 1, 2, and 3 with x2 – 5x + 1 as the new dividend.

X2 + 5x

X – 2

x3 + 3x2 – x + 1

x3 – 2x2

5x2 – x + 1

5x2–10x

9x + 1

x

Step 5: Repeat steps 1, 2, and 3 with 9x + 1 as the new dividend.

X2 + 5x + 9

x3 + 3x2 – x + 1

-(x3 – 2x2)

5x2 – x + 1

-(5x2–10x)

9x + 1

-(9x -18)

19

x

X – 2

 

Note: x3 + 3x2 – x + 1 is the dividend.

x – 2 is the divisor.

X2 + 5x + 9 is the quotient.

19 is the remainder.

We can combine the quotient, divisor and remainder to get the polynomial as follows:

x3 + 3x2 –x + 1 = (x – 2) x (x2 + 5x + 9) + 19

Polynomial (P) = Divisor (D) x Quotient (Q) + Remainder(R)

Solution 2

4x – 5

x2 + x – 1 4x3 – x2 + x – 5

-(4x3 + 4x2 – 4x)

  • 5x2 + 5x – 5

-(-5x2 – 5x + 5)

10x – 10

Note that in each of the examples above, the degree of Q = degree of P – degree of D and that the degree of R is one less than the degree of D. The degree of a remainder is one less than that of the divisor.

Class activity

  1. Find the quotients and remainders when P(x) = 6x3 + 4x2 – x + 5 is divided by:
  2. 3x2 + 2x +1
  3. x3 – 3x + 2
  4. A polynomial is divided by x + 1. The quotient is 2x – 3 and the remainder is 3, find the polynomial.

SUB-TOPIC 3

Remainder and factor theorem

The long division method used in the previous sub-topic helps us to determine, not only the quotient but also the remainder.

Consider the example 1 of the previous sub-topic:

Find the remainder when x3 + 3x2 –x + 1 is divided by x – 2.

X2 + 5x + 9

x3 + 3x2 – x + 1

-(x3 – 2x2)

5x2 – x + 1

-(5x2–10x)

9x + 1

-(9x -18)

19

x

X – 2

Hence when P(x) is divided by x – 2, the remainder is 19

Now find P(2).

P(2) = (2)3 + 3(2)2 – (2) + 1 = 8 + 12 – 2 + 1 = 19.

From the above, you would have observed that when p(x) is divided by x – a, the remainder is f(a). This forms the basis of Remainder theorem.

The Remainder theorem states that if a polynomial p(x) is divided by x – a, the remainder is p(a).

More generally, if p(x) is divided by ax + b, then the remainder is p(

A special case of the remainder theorem is when p(x) leaves no remainder when it is divided by x – a. when this happens, we say x – a is a factor of p(x).

This modified theorem is called factor theorem and it is states that if p(a) = 0 then x – a is a factor.

Examples

  1. Find the remainder when 2x2 – 5x + 6 is divided by x – 3.
  2. Determine the values of p and q if (x – 1) and (x + 2) are factor of 2x3 + px2 – x + q.

Solution:

  1. Let f(x) = 2x2 – 5x + 6

Let R be the remainder when f(x) is divided by x – 3

Then,

R = f(3)

F(3) = 2(3)2 – 5(3) + 6

= 2(9) – 15 + 6

= 18 – 15 + 6

= 9

  1. Let f(x) = 2x3 + px2 – x + q

If x – 1 is a factor of f(x), then

F(1) = 0

F(1) = 2(1)3 + p(1)2 – (1) + q

= 2 + p – 1 + q

= p + q + 2 – 1

= p + q + 1

p + q + 1 = 0 …. (1)

If x + 2 is a factor of f(x), then

F(-2) = 0

F(-2) = 2(-2)3 + p(-2)2 – (-2) + q

= 2(-8) + p(4) – (-2) + q

= -16 + 4p + 2 + q

= 4p + q -14

4p + q -14 = 0 … (2)

Solving simultaneously…

Subtract (1) from (2)

3p – 15 = 0

3p = 15

P = 5

Substitute the value of p into (2)

q = 14 – 4p

q = 14 – 4(5)

q = 14 – 20

q = -6

Hence p = 5, q = -6

Class activity

  1. Find the remainders without performing long division when
  2. x3 +5x2 – 3x + 1 is divided by x + 1
  3. 2x3 – 4x2 + x – 3 is divided by x + 2
  4. If (3x – 1) is a factor of the polynomial f(x) = 4x3 – 4x2 – x + p, find the value of the constant p.

SUB-TOPIC 4

Zero polynomials

Given a polynomial function f(x), the value x = a such that f(a) = 0 is called the zero of the polynomial.

A zero of the function f(x) is a root of the equation f(x) = 0.

To obtain the zeros of a polynomial f(x), set f(x) = 0 and solve the equation.

Examples:

  1. Find the zeros of the following polynomials:
  2. F(x) = x2 – 7x + 12
  3. G(x) = x2 – 16
  4. 2 is a factor of 32, find the values of the constants and state the zeros of

Solution 1:

  1. Set f(x) = 0

x2 – 7x + 12 = 0

x2 – 3x – 4x + 12 = 0

x(x – 3) – 4(x – 3) = 0

(x – 3)(x – 4) = 0

x = 3 or x = 4

Hence the zeros of f(x) are 3 and 4.

  1. Set g(x) = 0

x2 – 16 = 0

(x – 4)(x + 4) = 0

x = 4 or x = -4

Hence the zeros of g(x) are 4 and -4.

Solution 2:

If 2 is a factor, it means the zeros of 2 when substituted into will give the complete value of

from 2

2

 

(x + 2) (x – 3) = 0

x = -2 or x = 3

3(3)3 2

3(-2)3 2

⇒

From equation (i),

Substitute

Hence,

3232

To get the third factor of divide by 2

 

2 32

32

  • 2

2

0 0 0

The third factor is ⇒ hence,

More generally, an nth degree polynomial will have n zeros while an equation of degree n will have n roots.

Class activity

Given that a and b are the zeros of the polynomial f(x) = x2 – x – 6 with a˃b, and that g(x) = f(x + 2), find:

  1. g(a) + g(b)
  2. g(a) – g(b)
  3. g(a) x g(b)

PRACTICE QUESTIONS

  1. Given that P(x) = ax2 + bx + 1, P(1) = 6 and P(-1) = 2, determine the values of a and b.
  2. In the identity ax2 + bx + c = x2 – 2, find a, b, c.
  3. Given that f(x) = x3 + 3x2 + 3x + 1 and g(x) = x3 +- 3x2 + 3x -1, find (a) f(x) + g(x) (b) f(x) – g(x) (c) f(x) x g(x)
  4. Given that p(x) = ax2 + bx +1, p(1) = 6 and p(-1) = 2, determine the values of a and b.
  5. Find the quotients and remainders when P(x) = 6x3 + 4x2 – x + 5 is divided by:
  6. 3x2 + 2x +1
  7. x3 – 3x + 2
  8. A polynomial is divided by x + 1. The quotient is 2x – 3 and the remainder is 3, find the polynomial.

EVALUATION

  1. The polynomial has the same remainder when divided by (x+2) and (x-1). Find the value of constant q. (A) -11 (B) -9 (c) -3 (D) -1
  2. The polynomial has a remainder 20 when divided by (x – 2). Find the value of constant P. (A) 8 (B) 6 (C) -6 (D) -8
  3. Find the remainder when is divided by x-2 (A) 28 (B) – 28 (C) – 56 (D) 56
  4. The expression leaves a remainder of 6 when divided by x-k. The positive value of k is (A) 1 (B) 2 (C) 3 (D) 4
  5. Given f(x) = ax3 + 2x2 + bx + c, and f(0) = f(1) = 4 and f(-2) = 8, find the values of a, b and c.
  6. When is divided by (x-1), (x+2) and (x-3) the remainders are -14, 16 and 36 respectively. Find the values of the constants p, q and r and hence determine the quotient and remainder when f(x) is divided by x-4.

WEEK TWO

TOPIC: POLYNOMIALS 2

SUB-TOPICS:

  1. Roots of cubic equation: Sum of roots; Sum of products of two Roots; Products of roots.
  2. Graphs of polynomial function.

SUB-TOPIC 1

The general cubic equation takes the form: 3 (because if it becomes a quadratic equation).

3 …(1)

Dividing through by a,

3 …(2)

Let the roots of this equation 2 be. Then 3

= 0 …(3)

2

2– 2

Collect like terms

…(4)

By comparing coefficients of equations (2) and (4)

  1. The sum of roots
  2. The sum of product of roots
  3. The product of roots

Example 1:

The roots of a cubic equation are such that obtain the equation the roots of which are 2, 2, 2.

Solution:

The three ways of obtaining 2, 2 and 2 include.

(a). Expanding 2

22+2

= 2 22

2+22

⇒ 22+22

2+22 2

2+222

(b). Expanding 2 we have:

2222222

2

(c). 22222

The required equation is 322+2)222+2(22)]2 = 0

3 2

32

Example 2:

One of the roots of the cubic equation.

Find the:

  1. Sum of the two other roots;
  2. Product of the two other roots.

Hence or otherwise, find the other two roots.

Solution

Let α, β and γ be the roots of the equation such that γ = 5, then

Given that

a = 1, b = -9, c = 23, d = 15

hence:

… (1)

Also,

= 15

15

= 3 … (2)

From equation (1) we have

Substituting 4 – for

= 3

If α =1, then β =3 and

If α = 3, then β = 1

Hence:

α=1; β=3,γ=5.

Class activity

  1. The equation 32 has roots Find the equation whose roots are 3,3,3.
  2. Write down the cubic equation with solutions such that and

SUB-TOPIC 2

Graphs of polynomial function

The shape of a polynomial graph depends on the degree of that polynomial.

Polynomials of degree one

The straight line is the graphical representation of polynomials of degree one. The coefficient of x gives us a measure of the gradient or slope of the line.

If a ˃ 0, the straight line rises as y increases when x also increases. If a ˂ 0, the straight line falls as y decreases when x increases. If the graphs below, the points A and B on the straight line are called the x and y intercepts respectively.

y

Y = ax – b

a ˂ 0

B

y

Y = ax +b

a ˃ 0

B

A

x

A

x

Determining x- and y-intercept

To find the x-intercept, put y = 0 and solve for x in the equation. The x-intercept is identified as the zero of the corresponding polynomial.

To find the y-intercept, put x = 0 and solve for y.

From the knowledge of the intercepts, one can easily sketch the graph of a polynomial of degree 1.

Polynomial of degree two

The parabola is the graphical representation of polynomials of degree two. It has two shapes which depends on whether the coefficient of x2 is positive or negative.

Determining x- and y-intercept

To find the x intercept, put y = 0 and solve for x. the values of x for which y = 0 are the zeros of the polynomial.

To find y-intercept, put x = 0.

Turning points

The lowest point A on the curve in graph 1 is a turning point and it is called Minimum point.

The highest point B on the curve in graph 2 is also a turning point and it is called Maximum point.

Polynomials of degree three

The curve of polynomials of degree 3 is usually called cubical parabola and it has two shapes depending on whether a ˃ 0 or a ˂ 0.

Examples:

  1. Sketch y = 2x -1 by first finding the slopes and intercepts on the axes
  2. Sketch y = x2 +2x – 3 showing the intercepts and turning points.
  3. Sketch the curve represented by y = 12 + 4x -3x2 – x3.

Class activity

  1. Show that (2x-1) is a factor of the polynomial f(x) = 8x3 – 8x2 + 1 and find the quadratic factor.
  2. Sketch the graphs of the following: (i) y = -3x + 2. (ii) y = 8 – 2x – x2. (iii) y = x3 + 2x2 – 5x – 6.

PRACTICE QUESTIONS

  1. The expression px2 + qx +6 is divisible by x-3, and has a remainder of 20 when it is divided by x + 1. Find the values of p and q.
  2. When the polynomial f(x) = px3 + qx + r (where p, q and r are constants) is divided by (x + 3) and (x – 2), the remainders are -12 in each case. If (x + 1) is a factor of f(x), find: (i) f(x); (ii) the zeros of f(x).
  3. Given that x – 2 is a factor of 2x3 – x2 – 8x +4, find the other two factors.
  4. The equation 32 has roots Find the equation the roots of which are 3,3,3.
  5. Factorise completely 4x3 – 8x2y – 9xy2 + 18y3.

EVALUATION

  1. If the polynomial x3 + px2 + qx – 6 has a factor (x – 1) and leaves a remainder of -24 when divided by (x + 1):
  2. find the constants p and q.
  3. factorise the polynomial completely and find its zeros.
  4. Factorise 432 and 32 completely.
  5. Write down the cubic equation with solutions such that and
  6. The remainders when f(x) = x3 + ax2 + bx + c is divided by (x – 1), (x + 2) and (x – 2) are respectively 2, -1 and 15, find the quotient and remainder when f(x) is divided by (x + 1).
  7. If the polynomial f(x) = ax2 + 13x = b and g(x) = 4x2 + px + q are divided by x – 1, the remainders are 12 and 16 respectively. It they are divided by x – 2, the remainders are 40 and 20 respectively. Find the values of the constant a, b, p and q and hence determine the values of x for which f(x) = g(x).

WEEK THREE

TOPIC: PERMUTATIONS

SUB-TOPICS:

  1. Permutation (arrangement).
  2. Cyclic permutation.
  3. Arrangement of identical objects.
  4. Arrangement in which repetitions are allowed.

SUB-TOPIC 1

Permutation (Arrangement)

Suppose we are interested in the different arrangement of two people in a line. If these two people are labelled a and b then the problem is the same as finding the different arrangement of the letters a and b.

The number of arrangement will be two i.e. ab and ba.

Suppose there are three people in the line labelled a, b, and c. finding the different arrangements will be the same as finding the arrangements of the letters a, b and c.

c

b

a

c

b

b

a

c

a

b

a

c

c

a

b

bac

bca

abc

cab

cba

acb

From the above, we see there are six different arrangements.

We can get the number of different arrangement of an arbitrary n terms.

NNo. of different arrangementsFormula
111
222 x 1
363 x 2 x 1
4244 x 3 x 2 x 1
51205 x 4 x 3 x 2 x 1

Based on the above pattern, the number of different arrangement of n objects will be:

. This product can be written as n for short. n is read n-factorial.

The factorial of a positive integer is the product of all integers less than or equal to that given number.

I have four balls of different colours: Blue (B), Green (G), Red (R) and Yellow (Y). If I pick three of the balls, the following are the possible results of picking in order:

BGR BRG BGY BYG BRY BYR

GBR GRB GBY GYB GRY GYR

RBG RGB RBY RYB RGY RYG

YBG YGB YBR YRB YGR YRG

TOTAL = 24

Each of these arrangement is called a permutation. For the above, we obtained 24 permutations of four (4) colours taking three at a time. The way this is done is as follows:

The 1st ball could be any of the four balls available;

The 2nd ball could be any of the three colours remaining;

The 3rd ball could be any of the two colours remaining.

Thus, we have. This is called the arrangement of 4 balls taking 3 at a time.

If we have five colours to arrange, taking 5 at a time, we will obtain permutations.

We apply the basic counting principle that: ‘‘If an activity ‘A’ can be performed in m and another activity ‘B’ can be performed in , then, the two activities can be performed one after the other in

Permutation can therefore be defined as each of several possible ways in which a set or number of things can be ordered or arranged. It is also all possible arrangement of a collection of things where the order is important.

Suppose we are only interested in the number of ways, the first and second positions can be taken by 4 people in a race, assuming there is no tie.

The first position can be taken in four ways by any of the four athletes. The second position can be taken in 3 ways by any of the remaining 3 athletes.

So, the number of ways the first and the second positions can be taken by four people in a race is. This arrangement is called permutation of 4 people taking 2 at a time and is denoted by

 

Also, the number of permutations of 8 objects taking 3 at a time is denoted by

 

 

Thus, the general formula is

 

This is the permutation of n objects taking r at a time.

Examples

  1. Evaluate the following: (a) 7! (b) 0! (c) 1!
  2. Simplify
  3. Evaluate each of the following: (a) (b)
  4. Find the number of ways of arranging the letters of the word, EIGHT.

Solution:

  1. (a)

(b) 0! = 1

(c) 1! = 1

2.

  1. (a)

(b)

  1. EIGHT has five different numbers. Hence, the number of permutation is

 

Class activity

  1. Evaluate each of the following: a) b)
  2. In how many ways can five bulbs of different colours be arranged in five socket in a row?
  3. In how many ways can the letters of the word ENGLISH be arranged?

SUB-TOPIC 2

Cyclic permutation

In cyclic permutation, we are concerned about arrangement of this about a circular object.

If the letters A, B, C, D are arranged in that order in a circle, and then A is moved to B’s position and B to C’s position, C to D’s position, and D to A’s position, we obtain the same arrangement. i.e,

A

D

C

B

D

C

B

A

 

To obtain different arrangement, we fix one of the letters and arrange the remaining three in the remaining spaces. This gives 3! arrangements of the 4 letters.

A

D

C

B

A

C

D

B

A

D

B

C

A

C

B

D

A

B

D

C

A

B

C

D

In general, the number of ways of arranging ‘n’ objects in a circle is given by:

No. of ways = 1 x (n-1)!

When beads are threaded in a ‘ring’ the clockwise and the anticlockwise arrangements are not distinguishable and the ring can be turn over.

Thus, the number of distinct arrangements of ‘n’ objects round a circular ring which can be turned over is:

Examples:

  1. In how many ways can 8 boys be arranged at a round table?

Solution:

Total number of arrangements = 1 x (8-1)! = 7! = 5040

  1. Seven beads of different colours are threaded in a ring. How many different arrangement is possible?

Solution:

No. of arrangements =

Class activity

  1. In how many ways can eight boys be arranged around a circular table?
  2. A family of seven is to be seated round a table. In how many ways can this be done if the father and the mother are to sit together?
  3. In how many ways can eight men be seated at a round table if two particular men refused to sit together?

SUB-TOPIC 3

Arrangements of identical objects

Consider the arrangement of 8 bulbs (4 red, 3 blue and 1 yellow) in a row, there are 8! Possible permutations (arrangements). Out of these permutations, 4! Permutations involving changes in position of the red bulbs are not distinguishable and the 3! Permutations of the blue bulbs are also not distinguishable.

Thus, the number distinct permutations of the 8 bulbs (4 red, 3 blue and 1 yellow) =

In general, the number of distinct permutations of the ‘n’ objects containing p of one type of q of a second type and r of a third type is given by:

Example

In how many ways can the letters of the following words be arranged ?

  1. ABAKALIKI
  2. MATHEMATICS

SOLUTION

a) ABAKALIKI 9 letters; letter A appears 3 times, K twice, I twice, B&L once each.

No. of permutation =

b) MATHEMATICS = 11letters ( 2 M’s, 2 T’s, 2 A’s, & others once)

 

Class activity

  1. Find the number of ways the letters of the words FURTHER can be permuted.
  2. In how many ways can the letters of the word STRANGE be arranged so that the vowels occupy only the odd places?
  3. Find the number of arrangement in the letters of the word CONGRATULATIONS if the letter A must be placed next to each other. (Leave your answer in terms of factorial).

SUB-TOPIC 4

Arrangements in which repetitions are allowed

Examples

  1. How many numbers greater than 600 can be formed from the digits 2, 3, 4, 5, 6, 7?
  2. How many four-digits even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 if
  3. No repetitions are allowed
  4. Repetitions are allowed.

SOLUTION

(a) Without repetition: Numbers greater than 600 that can be formed from the option of 6 digits 2, 3, 4, 5, 6, 7 can be a 3-digit number, 4-digit number, 5-digit number or 6-digit number.

For a 3-digit number to be greater than 600, the first digit must be any of the 2 digits (6 or 7)

The 2nd digit, any of the remaining 5 digits ( i.e, after choosing one digit from 6 or 7)

The 3rd digit, any of the remaining 4 digits.

Therefore,

For 3-digits numbers we have

For 4-digits numbers, we have

For 5-digits numbers, we have

For 6-digits numbers, we have

Total = 40 + 360 + 720 + 720 = 1840

With repetition: the numbers of digits greater than 600 that would be formed with 2, 3, 4, 5, 6 and 7 will be:

For 3 digits numbers, there will be

For 4 digits numbers,

For 5 digits numbers,

For 6 digits numbers,

Therefore, total numbers that can be formed = 72+1296+7776+48656 = 57,800

(b) For an even number, the last digit must be 2 or 4 or 6.

Without repetition: The last digit can be any of the 3 digits.

The 1st digit can be any of the remaining 5 digits

The 2nd digit can be any of the remaining 4 digits

The 3rd digit can be any of the remaining 3 digits.

Therefore number required =

With repetition: the number required =

Class activity

  1. How many 4-digits numbers can be formed from the digits 0, 1, 2, 3, …9

If: (a) repetitions are allowed (b) the last digit must not be zero and repetitions are not allowed.

  1. How many 4-digit odd numbers can be formed with the digits 1, 2, 3 and 4 if: (a) repetition is allowed (b) repetition is not allowed.
  2. How many numbers less than 3000 can be formed from the digits 1, 2, 3, 8 and 9, if no digit is used more than once?

PRACTICE QUESTIONS

  1. Simplify. (a) 24 (b) 42 (c) 72 (d) 27
  2. Five students are lined up in a row. How many arrangement could be made if the position of the last boy remains unchanged? (a) 120 (b) 12 (c) 21 (d) 24
  3. Find the number of ways in which the letters of the word STATISTICS could be arranged. (a) 15120 (b) 5120 (c) 2020 (d) 1512
  4. Seven students were late to a class. In how many ways can they occupy: (a) three available vacant seats? (b) Nine available vacant seats?
  5. In how many ways are there of arranging 3 different jobs between 5 men where any man can only do one job?

EVALUATION

  1. In how many ways can 8 people be seated on a bench if only 3 seats are available?
  2. If the mathematics department of a particular college has 5 members of staff and they are to pose for a photograph by standing in a row. How many different arrangements are possible?
  3. Five people are to have a dinner jointly. In how many ways can they sit round a table if a couple must sit together?
  4. How many numbers greater than 4000 can be formed using some or all the digits 6, 5, 4, 3 and 2 without repetition? How many of these will be even?
  5. How many words can be formed from the letters of VALEDICTORY provided the letters A, E, I, O, Y are not to be separated at all?

WEEK FOUR

TOPIC: COMBINATION

SUB-TOPICS:

  1. Combination (selection).
  2. Conditional arrangements and selection.
  3. Probability problems involving arrangement and selection.

SUB-TOPIC 1

Combination (Selection)

In many situations, we make selection without regard to the order. If a committee of 4 members is to be formed from 7 members of staff of DLHS, the order in which the numbers of the given committee are selected is not important.

Combination is therefore a way of selecting items from a collection such that (unlike permutation) the order of selection does not matter.

In selecting three colours from 5 colours: (B, G, R, W, Y), BGR, BRG, GBR, RBG, RGB, are counted as 6 different arrangements (permutations), although they consist of the same 3 colours. The 6 permutations thus represent one combination. Thus, each combination of three objects yields 3! permutation.

Now, the number of the permutations of 5 colours taking 3 at a time, i.e,

The number of combinations of 5 colours taking 3 at a time, i.e,

In general,

Examples

  1. Out of the five science club members of a school, A, B, C, D and E, just three are to be chosen to represent the school in an exhibition. In how many ways can this be done?
  2. In how many ways can a committee of 3 chemistry teachers and 5 mathematics teachers be formed from 6 chemistry teachers and 10 mathematics teachers?

Solution

  1. The three representatives can be selected in
  2. The chemistry teacher can be selected in 6C3 ways and the mathematics teacher can be selected in 10C5 ways.

Total number of ways = 6C3 x 10C5

=

=

=

=

Class activity

  1. In how many ways can a disciplinary committee of 3 be formed from 10 members of staff of a college?
  2. A man travels from Nigeria to Ghana by air and from Ghana to Liberia by ship. He returns by the same means. He has 6 airlines and 4 shipping lines to choose from. In how many ways can he make his journey without using the same airline or shipping line twice?

SUB-TOPIC 2

Conditional arrangement and selection

When restrictions are placed on arrangements or selection, then, the permutation or combination is said to be conditional.

Example 1:

Find the number of ways 6 people can be seated in a round table if two particular friends must sit next to each other.

Solution:

If two people must seat next to each other, the number of ways these friends can sit is 2!

Therefore, the number of ways six people can sit in a round table with two friends that must be together is

Example 2:

A committee of 4 people is to be chosen from 5 married couples. Find how many ways the committee can be chosen if: (i) everyone is equally eligible; (ii) the committee should include at least one woman.

Solution:

i) 5 married couples includes 5 men and 5 women. Since everyone is equally eligible, then, the possible ways of selecting 4 people for the committee are:

4 men and 0 women or

3 men and 1 women or

2 men and 2 women or

1 man and 3 women or

0 men and 4 women.

i.e,

 

 

ii) If at least one woman must be in the committee, then, the possible ways of selecting 4 members of the committee from the couples (5men & 5women) are:

3 men and 1 woman or 2 men and 2 women or 1 man and 3 women or 4 women

i.e,

 

Class activity

  1. An excursion group of 4 is to be drawn from among 5 boys and 6 girls. Find the number of ways of choosing the excursion group if the group:
  2. is to be made up of an equal number of boys and girls;
  3. is to be either all boys or all girls;
  4. has no restrictions on its composition.
  5. A candidate is expected to attempt 12 out of 15 questions. In how many ways can this be done if:
  6. the candidate is to attempt any 12 question;
  7. the first 8 questions are compulsory;
  8. a question is outside the syllabus and hence cannot be completed.

SUB-TOPICS 3

Probability problems involving arrangement and selection

Example 1:

A box contains 10 red, 3 blue and 7 black balls. If three balls are drawn at random, what is the probability that: (a) all 3 are red, (b) all 3 are blue, (c) one of each colour is drawn?

Solution:

Number of ways of selecting any 3 balls from 20 balls = number of element in the sample space.

Number of ways of selecting 3 red balls out of 10 =

(a) P(all the 3 balls are red) =

(b) P(all 3 are blue) =

(c) P(1 red, 1 blue & 1 black) =

Example 2:

Three-digit numbers are formed from the digits 1, 2, 4, 5 and 6. If repetition is not allowed and a number is picked at random, find the probability that it is a multiple of 5 or an odd number.

Solution:

Since probability is involved, we find the sample space for 3-digit numbers formed from 5-digits i.e 5P3.

n(sample space) =

  1. To get multiple of 5, the last digit must be 5. That implies that

The 1st digit can be any of the remaining 4 digits.

The 2nd digit can be any of the remaining 3 digits

No. of ways = 1 x 4 x 3 = 12 ways.

Pr(multiple of 5) =

  1. To get odd number, the last digit will be any of 1 and 5

The first digit can be any of the remaining four

The second digit will be any of the remaining three.

No. of ways = 2 x 4 x 3 = 24ways.

Pr(odd number) =

Hence, pr(multiple of 5 or odd number) =

=

=

Class activity

  1. How many committee of size 5 consisting of three men and 2 women can be selected from 8 men and 6 women if a certain man must not be in the committee?
  2. A bag contains 5 white, 2 black and 3 green balls. If three ball are drawn at random, find the probability that:
  1. All three are green
  2. All three are white
  3. 2 are white and 1 is black
  4. At least, one is black
  5. 1 of each colour is drawn.

PRACTICE QUESTION

  1. How many committee of size 5 consisting of 3 men and 2 women can be selected from 8 men and 6 women if a certain man must not be in the committee? (a) 315 (b) 525 (c) 840 (d) 1287
  2. In how many ways can 9 bulbs be selected from 4 red, 5 green and 6 yellow bulbs if 3 of each colour are to be selected? (a) 800 (b) 120 (c) 40 (d) 27
  3. The number of ways of arranging 9 men and 8 women in a row, when the women occupy the even places is — (a) (b) (c)!8! (d)
  4. A panel consists of 5 men and 4 women. What is the probability of 4 men and 2 women?

(a) (b) (c) (d) .

5. Five digit numbers are formed from digits 4, 5, 6, 7 & 8

  1. How many of such numbers can be formed if repetition of digit is (i) allowed (ii) not allowed?
  2. How many of the numbers are odd if repetition of digits is not allowed?

EVALUATION

  1. Find the number of ways of arranging 9 men and 8 women in a row, if the women occupy the even places.
  2. If find the value of n.
  3. A panel of 5 jurist is to be chosen from a group of 6men and 7 women. Find the number of different panels that could be formed if: (a) a particular man must serve on the panel (b) there is no restriction.
  4. A business man intends to give a dinner party for 6 of his 10 friends. If 2 of them will not attend the party together, in hoe many ways can he select his guests?
  5. A family of 7 is to be seated round a table. In how many ways can this be done, if the father and the mother are to sit together?
  6. Four delegates are to be chosen from 8 members of staff of a college. If 2 of them are senior members of staff, how many different delegations are possible if: (i) only one of the senior members of staff must be in the delegation? (b) the two senior member of staff must be included?

WEEK FIVE

TOPIC: BINOMIAL EXPANSION 1

SUB-TOPICS:

  1. Pascal triangle.
  2. Binomial expansion of (a+b)n , where n can be positive integer, negative integer or fractional value.

SUB-TOPIC ONE

The Pascal’s triangle is a format for getting the coefficients of expansions. It applies to binomial and binomial formed from a reduced polynomial.

Consider the expansion of each of the following:

(a+b)0

(a+b)1

(a+b)2

(a+b)3

(a+b)4

(a+b)5

By ordinary expansion of algebraic terms, we have:

(a+b)0 = 1

(a+b)1 = 1a + 1b

(a+b)2 = 1a2+2ab+1b2

(a+b)3 = 1a3+3a2b+3ab2+1b3

(a+b)4 = 1a4+4a3b+6a2b2+4ab3+1b4

(a+b)5 = 1a5+5a4b+10a3b2+10a2b3+5ab4+1b5

Consider the array of coefficients of a and b. We can display it as follows:

 

n (power)

 

1 0

1 1 1

Row 1 2 1 2

1 3 3 1 3

1 4 6 4 1 4

1 5 10 10 5 1 5

We call the array of coefficients displayed above Pascal triangle named after the celebrated French Mathematician and Physicists Blaise Pascal (1623-1662) noted for his essay on conic section in 1640 and first invention of calculating machine in 1642.

Note the following:

  • A general binomial is of the form (a+b)n .
  • There are n + 1 terms.
  • The expansion is homogenous i.e.the sum of the powers of a and b in each term of the expansion is n.
  • As the power of a descends (starting from n till it reaches 0), the power of b ascends (starting from 0 till it reaches n) and vice versa.

Examples:

  1. Using Pascal’s triangle, expand

Solution:

Let n=4

1 4 6 4 1

Understand that the index (power) of descends as that of ascends.

 

  1. Use Pascal’s triangle to obtain the value of (1.025)4, correct to three decimal places.

Class activity

  1. .

SUB-TOPIC 2

Binomial Expansion of (a+b)n

Binomial Theorem

ncr

ncr =

ncr = ncr

The Binomial Theorem for a positive Integral Index

If a and b are any numbers and n is a positive integer, then

nC0an+nC1an-1b+nC2an-2b2+…+nCran-rbr+…nCnbn

Note:

  1. The number of terms in the expansion is n+1. That is one more than the index of binomial.
  2. The (r+1) term in the expansion of the binomial is called the general term and denoted by Tr+1 =nCran-rbr

The Binomial Theorem for Negative and Fraction

When n is not a positive integer, the expansion becomes

 

Provided a is numerically less than unit. That is -1<< 1. This means that the various coefficients cannot be expressed as nC0, nC1, nC2 etc because they have no meaning when n is not a positive integer.

Again, the theorem can be applied only when the first term of the binomial is unity. If not, the binomial must first be reduced to this form. For example, to expand (t+a)n has to be put in this form .

Examples:

4C0 + 4C1 + 4C2 + 4C3 + 4C4

=

  1. Expand to five terms.

Solution:

Recall:

n = -3,

  1. Expand to four terms.

Solution:

Note the first term of the binomial is not unity. Let’s reduce to the form

n= ½,

Class activity

  1. Determine the coefficient of from the expansion of .
  2. Expand to five terms.
  3. Expand to three terms.

PRACTICE QUESTIONS

  1. Using the binomial theorem, expand (1 + 2x)5, simplifying all the terms. Hence calculate the value of (1.02)5 correct to six significant figures.
  2. If the first three terms of the expansion of the expansion of (1+px)5 in ascending powers of x are 1+20x+160x2 find the values of n and p.
  3. (a) Write down the binomial expansion of (1+y)8, simplifying all the terms.

(b) Using the substitution y = x – x2 in (a), deduce the expansion of (1+ x – x2)8 in ascending powers of x as far as the term in x4.

(c) Find, by inspection, a value of x such that 1+ x – x2 = 1.09. Hence, evaluate (1.09)8 correct to three decimal places.

  1. Write down the binomial expansion of , simplifying all its coefficients.
  2. Obtain the first five terms of the expansion of

ASSIGNMENT

        1. The _______ triangle is a format for getting the coefficient of expansions. (a) Binomial (b) right-angled (c) Pascal’s (d) array.
        2. Pascal’s triangle was named after the French Mathematician and Physicists called_________ (a) Newton (b) Blaise Pascal (c) Cramer (d) Laplace
        3. The general binomial is of the form ______ (a) (a+b)n (b) (a+b)r (c) (a+b)! (d) aCb
        4. The (r+1) term in the expansion of the binomial is called the ________ term. (a) Simple (b) Difficult (c) Binomial (d) General.
        5. Expand
        6. Expand.

WEEK SIX

TOPIC: BINOMIAL EXPANSION 2

SUB-TOPICS:

  1. Finding nth term.
  2. Application of binomial expansion

SUB-TOPIC 1

Finding the nth term

The binomial expansion of is given as:

=

This expansion is true for any natural number value of n, large or small, but when n is a large natural number then is small. Indeed, the larger the value of n, the closer becomes to zero. The n notation for this,

 

the limit of as n→ is 0.

Also, as n→, the closer the expansion above becomes to the sum of terms:

Here the ellipsis (…) at the end of the expansion means that the expansion never ends, that is it has infinite number of terms.

Now we can use the sigma notation and write:

 

Notice the symbol for infinity (∞) at the top of the sigma, this denotes the fact that the sum is a sum of an infinite number of terms.

Examples:

  1. How many terms will there be in the expansion of. Find the fourth term.

Solution:

n=11, x=2x and

  1. Number of terms = n+1 = 11+1 =12
  2. From the formula,

11-3r = 2

-3r =2-11

-3r = -9; 3r = 9 ; hence r=3 (i.e. the 4th term)

 

  1. Find the term x2 and then term independent of x in the expansion of find the fifth term:

Solution:

  1. The general term is

When it is x2, then 12-2r = 2

r=5 (i.e. the sixth term)

Hence,

=

=

=

  1. If the term is to be independent of x, then 12-2r = 0

r =6 (i.e. the 7th term)

= = =

The independent term of x is

Class activity

  1. Find the term independent of x in the expansion
  2. How many terms will there be in the expansion?

SUB-TOPIC 2

Application of Binomial expansion in approximation

If x is very small, we can take 1 + nx as an approximation of is taken as 1-nx.

Examples:

Find the linear approximation of the following:

  1. (1.01)5 (b) (1.02)4 (c) (1.05)3 (d) (0.98)5

Solution: (a) (1.01)5

Expansion of (1+x)n = 1+nx= (1+0.01)5=1+5×0.01= 1+0.050= 1.050

  1. (1.02)4

Solution: Expansion of (1+x)n = 1+nx

(1.02)4 = (1+0.02)4≈1+4×0.02

= 1+0.080= 1.080

(c) (1.05)3; Expansion of (1+x)n = 1+nx

(1.05)3 = (1+0.05)4≈1+3×0.05= 1+0.15= 1.150

(d) (0.98)5

Solution: Expansion of 1+xn = 1+nx

(0.98)5 = (1-0.02)5≈1-5×0.02= 1-0.1= 0.09

EVALUATION

Find the linear approximation of the following:

  1. (1.002)15 (ii) (0.99)8 (iii) (2.004)7

PRACTICE QUESTIONS

  1. How many terms will there be in the expansion of
  2. Find the terms in the expansion of . Find the fift term.
  3. Expand and hence evaluate (0.98)7 correct to three decimal places.
  4. Find the linear approximation of the following (a) (2.004)7 (b) (1.003)11 (c) (d) .
  5. Using the binomial theorem, write down and simplify the first seven terms of the expansion of (1+2x)10 in ascending powers of x. use your expansion to show that 1.210˃6.19.

WEEKEND ACTIVITY:

  1. Find the number of terms in the expansion of . Find the third term.
  2. Find the term x2 and independent of x in the expansion .
  3. Find the coefficient of x-4 in the expansion.
  4. Determine the coefficient of from the expansion of .
  5. Expand to four terms.

WEEK SEVEN

MID TERM BREAK.

WEEK EIGHT

TOPIC: ROOTS OF QUADRATIC EQUATION 1

SUB-TOPICS:

  1. Quadratic equation (completing the square and formula method).
  2. Sum and product of roots of quadratic equation.
  3. Finding quadratic equation given sum and product of roots, x2 – (sum of roots) + (product) = 0.
  4. Condition for quadratic equation to have: (i) Equal roots (b2 = 4ac) (ii) Real roots b2 > 4ac (iii) No roots b2 < 4ac

SUB-TOPIC 1

Quadratic equation (completing the square and formula method).

A quadratic equation (trinomial) in one variable is a three termed equation in which the highest power of the variable is two. The general quadratic equation in variable x is of the form ax2 + bx + c = 0 where a ≠ 0.

In general, a quadratic equation has two solutions which may or may not b+e equal.

There are four major methods of solving quadratic equations. They are:

  • factorisation method;
  • completing the square method;
  • formula method;
  • graphical method.

Factorisation

Examples

Solve the following by factorisation

Solution

(a) Product of 1st and 3rd term = -8x2

Factor of -8x that will sum to -2x (the middle term) = -4x and +2x

Replacing the middle term with the two factors, we have:

(b) Product = -120x2

Sum = 14x

Factors = -6x and +20x

Completing the square method

Given ax2 + bx + c = 0, a ≠ 0

Write the equation as ax2 + bx = -c

Divide each term by the coefficient of x2. Find half of the new coefficient of x, square it and add this to both sides of the equality sign. This makes the expression on the LHS a perfect square.

Take the square root of both sides and determine the values of x to the desired accuracy.

The above is the process of solving quadratic equation using the method of completing the square.

Quadratic formula

The formula derived above is the quadratic formula.

Example:

Use the quadratic formula to solve

Solution:

a = 2; b = 7; c = -15

Class activity

  1. Use the three methods discussed above to solve the following quadratic equations:
  2. 2x2 – 17x – 9 = 0
  3. 3x2 + 10x – 12 = 0
  4. 8x2 + 34x + 21 = 0

SUB-TOPIC 2

Sum and product of roots of quadratic equation

Recall the completing the square method and general formula ‘ of the general equation

In this case the two roots of the equation are = or

It is possible to represent the two roots by then we have

 

 

 

 

 

 

Product of roots

 

From the above, if we have

We can rewrite this as .

Also, if α and β are the roots of the quadratic equation, then;

 

 

 

Therefore,

and

Given a quadratic equation, we can find the sum and product of the roots using the above information.

Example 1: if the roots of the equation are

Find

Solution: In the equation

 

The roots of the equation are

Then (a)

(b)

(c)

From (a)

 

(d)

 

 

24

= 40

Example 2:- if the roots of the equations

are (c) α2 – β2

Solution :- (i) The equation is

 

 

 

 

has no expression, then

 

 

 

But we want

 

 

 

 

 

This cannot be determined

The equation is

 

 

 

Following example I (c) we can express this as

(

 

 

 

Example 3:- Express the following in terms of

 

Solution:-

  1. .

 

 

.

 

 

Class activity

  1. are the roots of the equation

Find

 

 

  1. Given , prove that

 

 

SUB-TOPIC 3

Find quadratic equation given sum and product of roots

We have been able to establish that given α and β as the roots of a quadratic equation where a, b and c are constant and, then,

 

Also, and

Therefore, given the roots of a quadratic equation, the equation can be gotten this way:

Example 1:-

Given that the roots of equation are 3 and 7, find the equation.

Solution:

Roots of the equation are 3 and 7

Sum of roots

Product of roots (

 

 

Example 2:- Find the equation whose roots are -8 and 2.

Solution: Roots of the equation are -8, 2

 

 

The equation is

 

Example 3: If the roots of the equation are, find the equation.

Solution: Roots are

 

 

 

 

Multiply through by 12

 

Class activity

  1. Construct and simplify equations whose roots are given below:

 

  1. , write out the equation whose roots are . Find the values of from your equation.

SUB-TOPIC 4

Condition for quadratic equation to have:

 

In this section, we want to see some properties of roots. We can determine the type of roots that a particular equation will have.

The quadratic equation of

The part of the roots under square root sign is called the discriminant (i.e.) of the roots of the quadratic equation. This is because it can be used to determine the nature of the roots.

 

This means that whatever is contained inside the square root.

 

The quadratic is said to have coincident roots. This happens when the quadratic equation is a perfect square.

Example 1:- Consider the equation

SOLUTION: By factorization

 

 

 

 

(ii) Real roots: When b2 is greater than 4ac then the value under the roots sign is a positive number that is . The square root of such number will have two real values, one positive and other negative. The equation is said to have two distinct real roots.

Example 2: Solve the equation

 

,

 

 

 

(iii) Imaginary roots: When b2 is less than 4ac then the value under the root sign is a negative number i.e This shows that the root is not a real number. We say that the equation has imaginary roots.

Example 3: Find the roots of the equation

Solution: a=1 , b=2 c=5

Then

 

The roots are imaginary roots

Example 4:- Without solving the equation determine whether the equation has two different roots, coincident roots or imaginary roots.

Solution:- Rewrite the equation

 

Then a=5, b=-5, c=1

Then b2 = 25

 

 

Since

Then the equation has two real different roots.

Example 5:- Determine the nature of the roots equations without solving them.

 

Solution :

Rewrite the equation

 

To find the nature of the roots

 

 

 

 

The equation has two distinct real roots.

 

 

 

 

The equation has no real roots.

Class activity

Determine the nature of the roots of the following equations without necessarily solving them:

 

  1. Two real and different roots?
  2. Coincident and real roots
  3. No roots

PRACTICE QUESTIONS

        1. Given that are the roots of an equation such that the equation..
        2. If the equation has coincidental roots, find the value of P. (a)
        3. If
        4. If ( where p and q are constant, find the possible values of q.
        5. The roots of the equation are α and β, with α greater than β. Find the values of: (i) α – β (ii) α2 – β2.
        6. If the roots of the equation 4 Find the equation whose roots are

ASSIGNMENT

The roots of the equation are where find the value of k.

Find without necessarily solving the equation, the nature of the roots of the equation 3 . The equation (a) distinct two roots. (b) has no roots (c) coincident real roots (d) none of the above

If are the roots of the equation find (a)

Form the equation whose roots are -4 and 9.

The roots of the equation are, where m is a constant. If find the value of m.

If the sum of the squares of the roots of the equation is 1, show that

WEEK NINE

TOPIC: ROOTS OF QUADRATIC EQUATIONS 2

SUB-TOPICS:

  1. Quadratic functions (Simultaneous Equations: One linear, One quadratic)
  2. Solution of problems on roots of quadratic equation.
  3. Maximum and minimum values.

SUB-TOPIC 1

Quadratic functions (Simultaneous Equations: One linear, one quadratic)

We have discussed different ways but we need to mention that graphical solution is very important aspect of solving quadratic equations. This is because with graphical solution a lot of other problems can be solved.

The graph of the quadratic equation called parabola. Some call it cup or cap. The quadratic expression is equated to y and it is called a quadratic function. The example below show the graphical solution of quadratic function.

Example 1: Solve graphically, the equation 2

Solution:

Draw the table of values for the equation 2

01234
22712303122748
3210-1-2-3-4
-2-2-2-2-2-2-2-2
28124-2082242

Choose a convient scale, on and, on let represents and on represents

From the graph we find the point here the curve intersects

The graph is also useful to determine the minimum value of the minimum value of we have minimum point when and maximum point when

Simultaneous Equations

When solving simultaneous equation (you are already used to solving it graphically). In situation where one equation is linear and the second is quadratic, it can be solved by substitution as well as solving graphically.

In graphical solution of one linear-one quadratic simultaneous equation, there are three possible relationships between the straight line (linear) and the parabola (quadratic). They are:

  • Line intersecting with curve
  • Line touching curve at a point (tangent)
  • Line not intersecting the curve.

Example 2: Solve the simultaneous equations: 22

Solution: By substitution:

22

22

Since, then

Hence, adding (iii) together we get

From (ii) .

Example 3: Given the simultaneous equations:

2 and

Show on the graph the points of interest. Hence write out the values of .

Solution:

2

 

Table of values for 2 and

2

2362516941014

From the above, there was no intersect of the curve and the straight line. The solutions to the two equations cannot be determined because there is no point of intersection.

The points of intersection give the solution.

Example 3: On the same axes, plot the graph of y = 2x2-5 + 4 and y = 2x + 3. Hence find the points of intersection of the two graphs.

Solution:

Prepare the table of values for the functions given above.

y = 2x2-5 + 4

X-4-3-2-10123456
2x232188202818325072
-5x20151050-5-10-15-2025-30
+4+4+4+4+4+4+4+4+4+4+4+4
Y563722114127162946

Choose a convenient scale.

The points of intersections x = 0.2 and 3.3

The above example shows the case of the line intersecting with the curve.

Example 4: solve the simultaneous equation y = x2-2x + 2 and y = 4x -7. Interpret your result geometrically.

Solution:

Eliminate y to obtain: x2-2x + 2 = 4x -7 ⇒ x2 – 6x + 9 = 0

By factorisation:

(x – 3)(x – 3) = 0⇒x = 3(twice).

From y = 4x – 7 = 4(3) -7 = 5. The solution is x = 3 and y = 5.

Draw the graphs of the two equation to interpret it geometrically.

Table of values for y = x2 – 2x + 2. y = 4x – 7

x-2-101234x0234
x2410149164x081216
-2x820-2-4-6-8-7-7-7-7-7
+2+2+2+2+2+2+2+2y-7159
y145212510

 

The line y = 4x -7 intersects the curve y = x2 + 2x + 2 at only one point. Therefore, the solution to the equations is at the point x = 3 and y = 7.

Class activity

  1. Solve the simultaneous equations y = 4x – 1 and y =2x2 graphically and interpret your result geometrically.
  2. Solve for -1. Using a scale of 2cm to 1 unit on the x-axis and 2cm to represent 5 units on the y-axis.

SUB-TOPIC 2

Solutions of problems on roots of quadratic equation

Mathematics is important of life situation because of its application. You are used to problems leading to simple equations. We want to see the word problems leading to quadratic equations.

In order to solve such problems, you must take note of the following:

  1. Express the ideas involved in mathematical symbols.
  2. Write out the equation using the symbols.
  3. Solve the equation.
  4. Interprete your result.

Example 1: the product of two consecutive whole numbers is 506. Find the numbers.

Solution:

Let the numbers be x and (x + 1).

Then, x(x+1) = 506 ⇒ x2 + 1 = 506 (this is now quadratic equation)

X2 + x – 506 = 0

Solve by formula to find the values of x using the parameters below

a = 1 b = 1 c = -506

Example 2: There are two possible routes from Lagos to Ijebu Ode. One route is through Lagos/Ibadan express way which is 100km and the other is through Ikorodu-Epe covering a distance of 80km. A motorist going through express way can travel 10km per hour faster than the one going through Ikorodu and Epe and arrive Ijebu-Ode 5 minutes earlier as well. What is the time spent on the journey to Ijebu Ode by the motorist travelling through the express way?

Solution:

Let x be the speed of motorist going through Ikorodu/Epe and the speed of the one going through express way is x + 10.

Time taken by Ikorodu/Epe = 80/x.

Time taken by express way = 100/ (x + 10)

Hence, 80/x – 100/(x+10) = 1/12.

 

Form a quadratic equation from (i) above and solve it using formula and conclude.

Class activity

  1. The length of a rectangular field is 6m more than the width. If the area of the field is 72m2, find the dimensions of the field.
  2. Two consecutive odd integers are such that the sum of their reciprocals is Find the odd integers.

SUB-TOPIC 3

Maximum and Minimum values

The graph of as we have seen is a parabola. We have minimum point when and maximum point when

The maximum or minimum value (y) is

The curve is symmetrical about the line which is called the axis of symmetry.

If f(x) = 0, then,

      1. the curve cuts the horizontal axis if
      2. the curve touches the horizontal axis if
      3. the curve does not cut the horizontal axis if

Example 1:

Find the minimum value of and the corresponding the value of x for which y is a minimum.

Solution:

 

 

 

When x = -5/6, the expression in the brackets will be zero, hence the minimum is -49/12.

The corresponding value of x for which y is minimum is -5/6.

Note that x = -5/6 is the axis of symmetry of the parabola. Alternative, let the minimum value of y be ym then

 

Also the equation of the line of symmetry is

X = -b/2a = -5/6.

General evaluation:

  1. Solve the equations simultaneously and show the points of intersections

Y = 4 – 11x and y = 2x2-19

  1. Find the maximum value of y = 5 + 4x – x2 and the coordinates at the point where the curve y = 5 + 4x – x2, cuts the coordinates axes.
  2. The formula gives the sum of consecutive whole numbers. If
  3. A father got his first son at 31 years. If the product of their ages is 816. Find the ages of the father and his son.

 

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