# POLYNOMIALS (2)

**SUBJECT: FURTHER MATHEMATICS **

**CLASS: SSS 1**

**WEEK TWO**

**TOPIC: POLYNOMIALS 2**

**SUB-TOPICS:**

- Roots of cubic equation: Sum of roots; Sum of merchandise of two Roots; Merchandise of roots.
- 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)

- The sum of roots
- The sum of product of roots
- 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:

- Sum of the 2 different roots;
- 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**

- The equation
^{32}has roots Discover the equation whose roots are^{3},^{3},^{3}. - 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 x

^{2}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:

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

**Class exercise**

- Present that (2x-1) is an element of the polynomial f(x) = 8x
^{3}– 8x^{2}+ 1 and discover the quadratic issue. - Sketch the graphs of the next: (i) y = -3x + 2. (ii) y = 8 – 2x – x
^{2}. (iii) y = x^{3}+ 2x^{2}– 5x – 6.

**PRACTICE QUESTIONS**

- The expression px
^{2}+ 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. - When the polynomial f(x) = px
^{3}+ 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). - On condition that x – 2 is an element of 2x
^{3}– x^{2}– 8x +4, discover the opposite two elements. - The equation
^{32}has roots Discover the equation the roots of that are^{3},^{3},^{3}. - Factorise fully 4x
^{3}– 8x^{2}y – 9xy^{2}+ 18y^{3}.

**EVALUATION**

- If the polynomial x
^{3}+ px^{2}+ qx – 6 has an element (x – 1) and leaves a the rest of -24 when divided by (x + 1): - discover the constants p and q.
- factorise the polynomial fully and discover its zeros.
- Factorise
^{432}and^{32}fully. - Write down the cubic equation with options such that and
- The remainders when f(x) = x
^{3}+ ax^{2}+ 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). - If the polynomial f(x) = ax
^{2}+ 13x = b and g(x) = 4x^{2}+ 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).