POLYNOMIALS (2)

SUBJECT: FURTHER MATHEMATICS

CLASS: SSS 1

 

 

 

WEEK TWO

TOPIC: POLYNOMIALS 2

SUB-TOPICS:

1. Roots of cubic equation: Sum of roots; Sum of merchandise of two Roots; Merchandise of roots.
2. Graphs of polynomial operate.

SUB-TOPIC 1

The overall cubic equation takes the shape: ³ (as a result of if it turns into a quadratic equation).

³ …(1)

Dividing by by a,

³ …(2)

Let the roots of this equation 2 be. Then ³

= 0 …(3)

²

²– ²

Acquire like phrases

…(4)

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

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

Instance 1:

The roots of a cubic equation are such that get hold of the equation the roots of that are ², ², ².

Resolution:

The 3 ways of acquiring ², ² and ² embrace.

(a). Increasing ²

²²+²

= ^{2 22}

²+²²

⇒ ²²+²²

²+^{22 2}

²+²²²

(b). Expanding ² we’ve got:

^2222222

²

(c). ²²²²²

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

^{3 2}

³²

Instance 2:

One of many roots of the cubic equation.

Discover the:

1. Sum of the 2 different roots;
2. Product of the 2 different roots.

Therefore or in any other case, discover the opposite two roots.

Resolution

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

On condition that

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

therefore:

… (1)

Additionally,

= 15

15

= 3 … (2)

From equation (1) we’ve got

Substituting 4 – for

= 3

If α =1, then β =3 and

If α = 3, then β = 1

Therefore:

α=1; β=3,γ=5.

Class exercise

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

SUB-TOPIC 2

Graphs of polynomial operate

The form of a polynomial graph depends upon the diploma of that polynomial.

Polynomials of diploma one

The straight line is the graphical illustration of polynomials of diploma one. The coefficient of x provides us a measure of the gradient or slope of the road.

If a ˃ 0, the straight line rises as y will increase when x additionally will increase. If a ˂ 0, the straight line falls as y decreases when x will increase. If the graphs beneath, the factors A and B on the straight line are referred to as the x and y intercepts respectively.

y

Y = ax – b

a ˂ 0

B

y

Y = ax +b

a ˃ 0

B

A

x

A

x

Figuring out x- and y-intercept

To search out the x-intercept, put y = 0 and remedy for x within the equation. The x-intercept is recognized because the zero of the corresponding polynomial.

To search out the y-intercept, put x = 0 and remedy for y.

From the information of the intercepts, one can simply sketch the graph of a polynomial of diploma 1.

Polynomial of diploma two

The parabola is the graphical illustration of polynomials of diploma two. It has two shapes which depends upon whether or not the coefficient of x² is constructive or damaging.

Figuring out x- and y-intercept

To search out the x intercept, put y = 0 and remedy for x. the values of x for which y = 0 are the zeros of the polynomial.

To search out y-intercept, put x = 0.

Turning factors

The bottom level A on the curve in graph 1 is a turning level and it’s referred to as Minimal level.

The very best level B on the curve in graph 2 can also be a turning level and it’s referred to as Most level.

Polynomials of diploma three

The curve of polynomials of diploma 3 is normally referred to as cubical parabola and it has two shapes relying on whether or not a ˃ 0 or a ˂ 0.

Examples:

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

Class exercise

1. Present that (2x-1) is an element of the polynomial f(x) = 8x³ – 8x² + 1 and discover the quadratic issue.
2. Sketch the graphs of the next: (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 the rest of 20 when it’s divided by x + 1. Discover the values of p and q.
2. When the polynomial f(x) = px³ + qx + r (the place p, q and r are constants) is split by (x + 3) and (x – 2), the remainders are -12 in every case. If (x + 1) is an element of f(x), discover: (i) f(x); (ii) the zeros of f(x).
3. On condition that x – 2 is an element of 2x³ – x² – 8x +4, discover the opposite two elements.
4. The equation ³² has roots Discover the equation the roots of that are ³,³,³.
5. Factorise fully 4x³ – 8x²y – 9xy² + 18y³.

EVALUATION

1. If the polynomial x³ + px² + qx – 6 has an element (x – 1) and leaves a the rest of -24 when divided by (x + 1):
2. discover the constants p and q.
3. factorise the polynomial fully and discover its zeros.
4. Factorise ⁴³² and ³² fully.
5. Write down the cubic equation with options such that and
6. The remainders when f(x) = x³ + ax² + bx + c is split by (x – 1), (x + 2) and (x – 2) are respectively 2, -1 and 15, discover the quotient and the rest when f(x) is split 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’re divided by x – 2, the remainders are 40 and 20 respectively. Discover the values of the fixed a, b, p and q and therefore decide the values of x for which f(x) = g(x).