**1.** (i) The graph does not meets x-axis at all. Hence, it does not have any zero.

(ii) Graph meets x-axis 1 time. It means this polynomial has 1 zero.

(iii) Graph meets x-axis 3 times. Therefore, it has 3 zeroes.

(iv) Graph meets x-axis 2 times. Therefore, it has 2 zeroes.

(v) Graph meets x-axis 4 times. It means it has 4 zeroes.

(vi) Graph meets x-axis 3 times. It means it has 3 zeroes.

**2.** (i) ** **x^{2 }– 2x – 8

Comparing given polynomial with general form *ax ^{2}+bx+c,*

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

We have, x^{2 }– 2x – 8

= x^{2 }– 4x + 2x – 8

*= x(x – 4) + 2 *(x – 4) = *(x – 4)(x + *2)

Equating this equal to 0 will find values of 2 zeroes of this polynomial.

(x-4)(x+2) = 0

*⇒ x *= 4, -2 are two zeroes.

(ii) 4s^{2 }– 4s + 1

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

We have, 4s^{2 }– 4s + 1

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

= (2s – 1)(2s – 1)

Equating this equal to 0 will find values of 2 zeroes of this polynomial.

*⇒ *(2s – 1)(2s – 1) = 0

Therefore, two zeroes of this polynomial are

(iii) 6x^{2 }– 3 – *7x*

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

We have, 6x^{2 }– 3 – *7x*

*= *6x^{2 }– *7x – 3 *= 6x^{2 }– 9x + 2x – 3

= 3x (2x – 3) +1(2x – 3) = (2x – 3)(3x + 1)

Equating this equal to 0 will find values of 2 zeroes of this polynomial.

*⇒ *(2x-3)(3x+1) = 0

Therefore, two zeroes of this polynomial are,

(iv) 4u^{2}+8u

Here, a = 4, b = 8 and c = 0

4u^{2}+8u = 4u(u+2)

Equating this equal to 0 will find values of 2 zeroes of this polynomial.

*⇒ *4u(u+2) = 0

*⇒ u = *0,-2

Therefore, two zeroes of this polynomial are 0, -2

(v) t^{2 }– 15

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

We have, t^{2 }– 15 *⇒ *t^{2}= 15*⇒ *t = ±√15

Therefore, two zeroes of this polynomial are √15, -√15

(vi) 3x^{2 }– x – 4

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

We have, 3x^{2 }– x – 4 = 3x^{2 }– 4x + 3x – 4

= x(3x – 4) +1(3x – 4) = (3x – 4) (x+1)

Equating this equal to 0 will find values of 2 zeroes of this polynomial.

*⇒ *(3x-4)(x+1) = 0

Therefore, two zeroes of this polynomial are

Let quadratic polynomial be *ax ^{2}+bx+c*

Let *a *and β are two zeroes of above quadratic polynomial.

∴ Quadratic polynomial which satisfies above conditions = 4x^{2 }– x – 4

Let quadratic polynomial be *ax ^{2}+bx+c*

Let *a *and β* *be two zeros of above quadratic polynomial.

∴ Quadratic polynomial which satisfies above conditions = 3x^{2 }– 3√2x +1

(iii) 0, √5

Let quadratic polynomial be *ax ^{2}+bx+c*

Let *a *and β be two zeros of above quadratic polynomial.

∴ a = 1, b = 0, c = √5

∴ Quadratic polynomial which satisfies above conditions = x^{2}+ √5

(iv) 1, 1

Let quadratic polynomial be *ax ^{2}+bx+c*

Let *a *and β* *be two zeros of above quadratic polynomial.

∴ a = 1, b = -1, c = 1

∴ Quadratic polynomial which satisfies above conditions = x^{2 }– x + 1

Let quadratic polynomial be *ax ^{2 }+ bx + c*

Let *a *and β be two zeros of above quadratic polynomial.

∴ a = 4, b =1, c =1

∴ Quadratic polynomial which satisfies above conditions = 4x^{2 }+ x + 1

(vi) 4, 1

Let quadratic polynomial be *ax ^{2}+bx+c*

Let *a *and β be two zeros of above quadratic polynomial.

∴ Quadratic polynomial which satisfies above conditions = x^{2 }– 4x + 1

**4.** (i)

Therefore, quotient = x – 3 and Remainder = 7x – 9

(ii)

Therefore, quotient = x^{2} + *x *– 3 and, Remainder = 8

(iii)

Therefore, quotient = -x^{2} – 2 and, Remainder = -5x + 10

**5.** (i)

∵ Remainder = 0

Hence first polynomial is a factor of second polynomial.

(ii)

∵ Remainder = 0

Hence first polynomial is a factor of second polynomial.

(iii)

∵ Remainder ≠ 0

Hence first polynomial is not factor of second polynomial.

Applying Division Algorithm to find more factors we get:

We have p(x) = g(x) x q(x).

Therefore, other two zeroes of (3x^{4 }+ 6x^{3 }– 2x^{2 }– 10x -5) are -1 and -1.

**7.** Let p(x) = x^{3 }– 3x^{2}+x+2, q(x) = (x – 2) and r(x) = (-2x+4)

According to Polynomial Division Algorithm, we have

p(x) = g(x).q(x) + r(x) *⇒ *x^{3}-3x^{2}+x+2 = *g(x).(x-2)-2x+4*

*⇒ *x^{3}-3x^{2}+x+2+2x-4 = *g(x). *(x-2) *⇒*x^{3}-3x^{2}+3x-2 = *g(x). *(x-2)

**8.** (i)

So, we can see in this example that deg p(x) = deg q(x) = 2

(ii)

We can see in this example that deg q(x) = deg r(x) = 1

(iii) Let p(x) = x^{2}+5x-3, g(x) = x+3

We can see in this example that deg r(x) = 0

**9.** Comparing the given polynomial with *ax ^{3} + bx^{2} + cx + d , *we get

*a=2,b=1,c=-5*and

*d*= 2.

(ii) Comparing the given polynomial with *ax ^{3} + bx^{2} + cx + d , *we get

*a =1, b = -4, c = *5 and *d *= -2.

p(2) = 2(2^{3}-4(2^{2}+5(2)-2 = 8-16+10-2 =0

p(1)=(1)^{3}-4(1)^{2}+5(1)-2 = 1-4+5-2 =0

∴ 2, 1 and 1 are the zeroes of *x ^{3} – *4x

^{2}+5x – 2.

**10.** Let the cubic polynomial be *ax ^{3} + bx^{2} + cx + d *and its zeroes be

*a, β*and y.

Here, *a =1, b = *-2, c *= *-7 and *d = 14*

Hence, cubic polynomial will be *x ^{3} – *2x

^{2}–

*7x +14.*

**11.** Since *(a – b) , a ,(a + b) *are the zeroes of the polynomial *x ^{3}* -3x

^{2}+ 3x +1.

**12.** Since 2 ± √3 are two zeroes of the polynomial *p(x) = *x^{4} – 6x^{3} – 26x^{2} +138x – 35.

Let *x* = 2 ± √3 ⇒ *x* – 2 =±√3

Squaring both sides, ⇒ x^{2} – 4x+ 4 =3 ⇒ x^{2} – 4x +1 = 0

Now we divide *p (x) *by x^{2} – 4x +1 to obtain other zeroes.

*∴ p(x) *= x^{4} -6x^{3} -26x^{2} +138x-35

= (x^{2} -4x+1)(x^{2} -2x-35) = (x^{2} -4x+1)(x^{2} -7x+5x-35)

= (x^{2} -4x+1)[x(x-7)+5(x-7)1 = (x^{2} -4x+1)(x+5)(x-7)

⇒ (x + 5) and (x-7) are the other factors of *p(x). *

∴ -5 and 7 are other zeroes of the given polynomial.

**13.** Let us divide x^{4} — 6.x^{3} +16x^{2} — 25x +10 by x^{2} — 2x + *k.*

∴ Remainder = (2k *—9)x—(8—k)k +10*

On comparing this remainder with given remainder, i.e. *x+ a, *

2k – 9 = 1 ⇒ 2k =10 ⇒ *k = *5

And *-(8 – k)k *+ 10 *=a ⇒** a = -(8 – 5) 5 + 10= -5*