### Chapter 2: Polynomials

Q
##### Polynomials CBSE NCERT Solutions

Question:

Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

(i)  $x^{2}-2x-8$               (ii) $4s^{2}-4s+1$             (iii) $6x^{2}-3-7x$

(iv) $4u^{2}+8u$                 (v)  $t^{2} -15$                       (vi) $3x^{2} -x -4$

(i)  $x^{2}-2x-8$

Let P(x) = $x^{2}-2x-8$

Zeroes of the polynomial are the value of  where P(x) = 0.

Putting P(x) = 0 then we get

$x^{2}-2x-8$

We can find zeroes of the polynomial using splitting the middle term method.

In splitting the middle term method we need to find two numbers where the sum is -2 and product is -8 x 1 = -8.

Therefore, we can write,

$x^{2}-4x+2x-8=0$

$\Rightarrow x(x-4)+2(x-4)=0$

$\Rightarrow (x-4)(x+2)=0$

Either $(x-4)=0 \quad or$  $(x+2)=0$

So,  x = 4 and x = -2.

Let $\alpha$ = 4 and $\beta$  = -2 are the zeroes of the polynomial.

$P(x) = x^{2} - 2x - 8 = 1 x^{2} + (-2)x + (-8)$

If we compare $P(x)$  with  $ax^{2} + bx +c$ then we get

a = 1, b = -2 and c = -8

Now we verify the relationship between the zeroes and the coefficients in two different cases.

Case I:

$Sum \quad of \quad zeroes = - \frac{coefficient of x}{coefficient of x^{2}}$

$\Rightarrow \alpha + \beta = - \frac{b}{a}$

LHS = $\alpha + \beta = 4 - 2 = 2$

RHS =  $-\frac{b}{a}$ $= -\frac{(-2)}{1}$$= 2$

$\therefore$ LHS = RHS

Case II:

$Product \quad of \quad zeroes = \frac{constant \quad term}{coefficient of x^{2}}$

$\Rightarrow \alpha \times \beta = \frac{c}{a}$

LHS =  $\alpha \times \beta$  = 4 x (-2) = - 8

RHS = $\frac{c}{a} =$  $\frac{-8}{1} = -8$

$\therefore$ LHS = RHS

In both cases, LHS is equal to RHS.

Therefore, the relationship between the zeroes and the coefficients are verified.

(ii) $4s^{2}-4s+1$

Let P(s) = $4s^{2}-4s+1$

Zeroes of the polynomial are the value of  where P(s) = 0.

Putting P(s) = 0 then we get

$4s^{2}-4s+1$

We can find zeroes of the polynomial using splitting the middle term method.

In splitting the middle term method we need to find two numbers where the sum is -4 and product is 4  1 = 4.

Therefore, we can write,

$4s^{2}-2s-2s+1=0$

$\Rightarrow 2s(2s-1)-1(2s-1))=0$

$\Rightarrow (2s-1)(2s-1))=0$

$(2s-1))=0 \Rightarrow s = \frac{1}{2}$

Let $\alpha = \frac{1}{2}$  and  $\beta = \frac{1}{2}$  are the zeroes of the polynomial.

P(s ) =  $4s^{2} - 4s + 1 = 4s^{2} + (-4)s + 1$

If we compare P(s) with  $as^{2} + bs + c$ then we get

a = 4, b = -4 and c = 1

Now we verify the relationship between the zeroes and the coefficients in two different cases.

Case I:

$Sum \quad of \quad zeroes = - \frac{coefficient of x}{coefficient of x^{2}}$

$\Rightarrow \alpha + \beta = - \frac{b}{a}$

LHS = $\alpha + \beta = \frac{1}{2} + \frac{1}{2} = 1$

RHS = $-\frac{b}{a} = - \frac{(-4)}{4} = 1$

$\therefore$ LHS = RHS

Case II:

$Product \quad of \quad zeroes = \frac{constant \quad term}{coefficient of x^{2}}$

$\Rightarrow \alpha \times \beta = \frac{c}{a}$

LHS = $\alpha \times \beta = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$

RHS = $\frac{c}{a} = \frac{1}{4}$

$\therefore$ LHS = RHS

In both cases, LHS is equal to RHS.

Therefore, the relationship between the zeroes and the coefficients are verified.

(iii) $6x^{2}-3-7x$

Let P(x ) = $6x^{2}-3-7x$ $= 6x^{2}-7x-3$

Zeroes of the polynomial are the value of x where P(x ) = 0.

Putting P(x) = 0 then we get

$6x^{2}-7x-3 = 0$

We can find zeroes of the polynomial using splitting the middle term method.

In splitting the middle term method we need to find two numbers where the sum is -7 and product is 6  x (-3) = -18.

Therefore, we can write,

$6x^{2}-9x+2x-3 = 0$

$\Rightarrow 3x(2x-3)+1(2x-3 )= 0$

$\Rightarrow (2x-3)(3x+1 )= 0$

Either $(2x-3)= 0 \quad or \quad (3x+1 )= 0$

So, x = 3/2 and x = -1/3

Let $\alpha = \frac{3}{2}$     and     $\beta = -\frac{1}{3}$   are the zeroes of the polynomial.

P(x) = $6x^{2}-7x-3$  = $6x^{2} +(-7)x+(-3)$

If we compare p(x) with  $ax^{2} +bx+c$ then we get

a = 6, b = -7 and c = -3

Now we verify the relationship between the zeroes and the coefficients in two different cases.

Case I:

$Sum \quad of \quad zeroes = - \frac{coefficient of x}{coefficient of x^{2}}$

$\Rightarrow \alpha + \beta = - \frac{b}{a}$

LHS = $\alpha + \beta = \frac{3}{2} - \frac{1}{3} = \frac{9-2}{6} = \frac{7}{6}$

RHS = $-\frac{b}{a} = - \frac{(-7)}{6} = \frac{7}{6}$

$\therefore$ LHS = RHS

Case II:

$Product \quad of \quad zeroes = \frac{constant \quad term}{coefficient of x^{2}}$

$\Rightarrow \alpha \times \beta = \frac{c}{a}$

LHS =  $\alpha \times \beta = \frac{3}{2} \times \frac{-1}{3} = \frac{-1}{2}$

RHS =  $\frac{c}{a} = \frac{-3}{6} = \frac{-1}{2}$

$\therefore$ LHS = RHS

In both cases, LHS is equal to RHS.

Therefore, the relationship between the zeroes and the coefficients are verified.

(iv) $4u^{2}+8u$

Let P(u) = $4u^{2}+8u$

Zeroes of the polynomial are the value of  u where P(u) = 0.

Putting P(u) = 0 then we get

$4u^{2}+8u=0$

We can find zeroes of the polynomial using common factor method.

4u is the common factor in the P(u) = $4u^{2}+8u$

Therefore, we can write,

$\Rightarrow 4u(u+2)=0$

Either 4u  = 0 or  (u+2) = 0

So, u = 0 and u = -2.

Let $\alpha$  = 0 and $\beta$  = -2 are the zeroes of the polynomial.

P(u) = $4u^{2}+8u = 4u^{2}+8u+0$

If we compare p(u) with $au^{2}+bu+c$ then we get

a = 4, b = 8 and c = 0

Now we verify the relationship between the zeroes and the coefficients in two different cases.

Case I:

$Sum \quad of \quad zeroes = - \frac{coefficient of u}{coefficient of u^{2}}$

$\Rightarrow \alpha + \beta = - \frac{b}{a}$

LHS = $\alpha + \beta = 0 - 2 = -2$

RHS = $-\frac{b}{a} = - \frac{8}{4} = -2$

$\therefore$ LHS = RHS

Case II:

$Product \quad of \quad zeroes = \frac{constant \quad term}{coefficient of u^{2}}$

$\Rightarrow \alpha \times \beta = \frac{c}{a}$

LHS =  $\alpha \times \beta = 0 \times (-2) = 0$

RHS =  $\frac{c}{a} = \frac{0}{4} = 0$

$\therefore$ LHS = RHS

In both cases, LHS is equal to RHS.

Therefore, the relationship between the zeroes and the coefficients are verified.

(v) $t^{2}-15$

Let P(t) = $t^{2}-15$

Zeroes of the polynomial are the value of  t where P(t) = 0.

Putting P(t) = 0 then we get

$t^{2}-15=0$

$\Rightarrow t^{2} = 15$

$\Rightarrow t = \pm \sqrt{15}$

So, $t = + \sqrt{15}$  and  $t = - \sqrt{15}$

Let $\alpha$  $\sqrt{15}$ and $\beta$  = $- \sqrt{15}$ are the zeroes of the polynomial.

P(t) = $t^{2}-15$ = $1t^{2}+0\times t + (-15)$

If we compare p(t) with $at^{2}+bt+c$  then we get

a = 1, b = 0 and c = -15

Now we verify the relationship between the zeroes and the coefficients in two different cases.

Case I:

$Sum \quad of \quad zeroes = - \frac{coefficient \quad of \quad t}{coefficient \quad of \quad t^{2}}$

$\Rightarrow \alpha + \beta = - \frac{b}{a}$

LHS = $\alpha + \beta = \sqrt{15} - \sqrt{15} = 0$

RHS = $-\frac{b}{a} = - \frac{0}{1} = 0$

$\therefore$ LHS = RHS

Case II:

$Product \quad of \quad zeroes = \frac{constant \quad term}{coefficient of t^{2}}$

$\Rightarrow \alpha \times \beta = \frac{c}{a}$

LHS =  $\alpha \times \beta = \sqrt{15} \times (-\sqrt{15}) = -15$

RHS =  $\frac{c}{a} = \frac{-15}{1} = -15$

$\therefore$ LHS = RHS

In both cases, LHS is equal to RHS.

Therefore, the relationship between the zeroes and the coefficients are verified.

(vi) $3x^{2} -x -4$

Let P(x) = $3x^{2} -x -4$

Zeroes of the polynomial are the value of  where P(x) = 0.

Putting P(x) = 0 then we get

$3x^{2} -x -4$

We can find zeroes of the polynomial using splitting the middle term method.

In splitting the middle term method we need to find two numbers where the sum is -1 and product is 3 x (-4) = -12.

Therefore, we can write,

$3x^{2}-4x+3x-4=0$

$\Rightarrow x(3x-4)+1(3x-4)=0$

$\Rightarrow (3x-4)(x+1)=0$

Either $(3x-4)=0 \quad or$  $(x+1)=0$

So,  x = 4/3 and x = -1.

Let $\alpha$ = 4/3 and $\beta$  = -1 are the zeroes of the polynomial.

$P(x) = 3x^{2} - x - 4 = 3x^{2} + (-1)x + (-4)$

If we compare $P(x)$  with  $ax^{2} + bx +c$ then we get

a = 3, b = -1 and c = -4

Now we verify the relationship between the zeroes and the coefficients in two different cases.

Case I:

$Sum \quad of \quad zeroes = - \frac{coefficient of x}{coefficient of x^{2}}$

$\Rightarrow \alpha + \beta = - \frac{b}{a}$

LHS = $\alpha + \beta = \frac{4}{3} - 1 = \frac{4-3}{3} = \frac{1}{3}$

RHS =  $-\frac{b}{a}$ $= -\frac{(-1)}{3}$ $= \frac{1}{3}$

$\therefore$ LHS = RHS

Case II:

$Product \quad of \quad zeroes = \frac{constant \quad term}{coefficient of x^{2}}$

$\Rightarrow \alpha \times \beta = \frac{c}{a}$

LHS =  $\alpha \times \beta$  $= \frac{4}{3} \times (-1)$ $= -\frac{4}{3}$

RHS = $\frac{c}{a} =$  $\frac{-4}{3}$

$\therefore$ LHS = RHS

In both cases, LHS is equal to RHS.

Therefore, the relationship between the zeroes and the coefficients are verified.

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