POLYNOMIALS

SUBJECT: FURTHER MATHEMATICS

CLASS: SSS 1

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:

a_nxⁿ + a_n-1x^n-1 + … + a₂x² + a₁x + a₀

where the highest power of the variable n is the degree of the polynomial; the numerical constants a_n, a_{n-1, …} a_2, 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) = a_nxⁿ + a_n-1x^n-1 + … + a₂x² + a₁x + a₀ and

Q(x) = b_nxⁿ + b_n-1x^n-1 + … + b₂x² + b₁x + b₀

are said to be equal if:

a_n = b_n

a_n-1 = b_n-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) = a_nxⁿ + a_n-1x^n-1 + … + a₂x² + a₁x + a₀ and

Q(x) = b_nxⁿ + b_n-1x^n-1 + … + b₂x² + b₁x + b₀

Then,

P(x) + Q(x) = (a_n + b_n) xⁿ + (a_n-1 + b_n-1) x^n-1 + … + (a₂ + b₂) x² + (a₁ + b₁) x + a₀ + b₀

Similarly,

P(x) + Q(x) = (a_n – b_n) xⁿ + (a_n-1 – b_n-1) x^n-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 (3^rd degree), by x-2 (1^st 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.