*NCERT Solutions of Chapter: Polynomials. NCERT Solutions along with worksheets and notes for Class 10.*

**Exercise 2.1**

**(1)** The graphs of y = p(x) are given in fig below, for some polynomials p(x). Find the number of zeroes of p(x), in each case.

(i)

(ii)

(iii)

(iv)

(v)

(vi)

**Ans-**

**(i)** Number of zeroes for p(x) = 0

Curve does not intersect the x- axis.

**(ii)** Number of zeroes for p(x) = 1

Curve intersects the x- axis at one point only.

**(iii)** Number of zeroes for p(x) = 3

Curve intersects the x- axis at three points.

**(iv)** Number of zeroes for p(x) = 2

Curve intersects the x- axis at two points.

**(v)** Number of zeroes for p(x) = 4

Curve intersects the x- axis at four points.

**(vi)** Number of zeroes for p(x) = 3

Curve intersects the x- axis at one point and touches at two points.

**Exercise 2.2**

**(1)** 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

**Ans –**

**(i)** x^{2} – 2x – 8

= x^{2} – 4x + 2x -8

=x(x – 4) +2(x – 4)

=(x – 4)(x + 2)

Therefore, the value of zeroes of polynomial x^{2} – 2x – 8 is

x – 4 = 0, ie, x = 4

x + 2 + 0, ie, x = -2

The zeroes of polynomial x^{2} – 2x – 8 are 4 and – 2.

Now,

Sum of the zeroes = 4 + (-2) = 2

Sum of the coefficient = Sum of the zeroes

Product of the zeroes = (4)(-2) = -8

Product of the coefficient = Product of the zeroes

Hence, relationship between the zeroes and the coefficients is verified.

**(ii)** 4s^{2} – 4s + 1

= 4s^{2} – 2s -2s -1

= 2s(2s – 1) +1(2s – 1)

=(2s – 1)(2s – 1)

Therefore, the value of zeroes of polynomial 4s^{2} – 4s + 1 is

Product of the coefficient = Product of the zeroes

Hence, relationship between the zeroes and the coefficients is verified.

**(iii)** 6x^{2} – 3 – 7x

6x^{2} – 7x – 3

= 6x^{2} – 9x + 2x -3

= 3x(2x – 3)+ 1(2x – 3)

= (2x – 3)(3x + 1)

Therefore, the value of zeroes of polynomial 6x^{2} – 7x – 3 is

Product of the coefficient = Product of the zeroes

Hence, relationship between the zeroes and the coefficients is verified.

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

= 4u (u + 2)

Therefore, the value of zeroes of polynomial 4u^{2} + 8u is

4u = 0, ie, u = 0

u + 2 = 0, ie, u = – 2

The zeroes of polynomial 4u^{2} + 8u are 0 and – 2.

Now,

Sum of the zeroes = 0 + -2 = -2

Sum of the coefficient = Sum of the zeroes

Product of the zeroes = (0)(-2) = 0

Product of the coefficient = Product of the zeroes

Hence, relationship between the zeroes and the coefficients is verified.

**(v)** t^{2} – 15

= t^{2} +225 – 30t

= t^{2} – 30t +225

= t^{2} – 15 t – 15t + 225

=t(t – 15) – 15(t – 15)

=(t – 15) (t – 15)

Therefore, the value of zeroes of polynomial t^{2} – 15 is

t – 15= 0, ie, t = 15

t – 15= 0, ie, t = 15

The zeroes of polynomial t^{2} – 30t +225 are 15 and 15.

Now,

Sum of the zeroes = 15 + 15 = 30

Sum of the coefficient = Sum of the zeroes

Product of the zeroes = (15)(15) = 225

Product of the coefficient = Product of the zeroes

Hence, relationship between the zeroes and the coefficients is verified.

**(vi)** 3x^{2} – x – 4

= 3x^{2} + 3x – 4x – 4

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

=(x + 1) (3x – 4)

Therefore, the value of zeroes of polynomial 3x^{2} – x – 4is

x + 1= 0, ie, x = -1

Product of the coefficient = Product of the zeroes

Hence, relationship between the zeroes and the coefficients is verified.

**(2)** Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.

**Ans-**

**Exercise 2.3**

**(1)** Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following:

**(i)** p(x) = x^{3} – 3x^{2} + 5x -3, g(x) = x^{2} – 2

**(ii)** p(x) = x^{4} – 3x^{2} + 4x + 5, g(x) = x^{2} + 1 –x

**(iii)** p(x) = x^{4} – 5x + 6, g(x) = 2 – x^{2}

**Ans-**

**(i)**

Dividend p(x) = x^{3} – 3x^{2} + 5x -3

Divisor g(x) = x^{2} – 2

Since degree (3) = 0 < 2 = degree (x^{2} – 2)

Therefore, quotient = x – 3, remainder = 7x – 9

**(ii)** p(x) = x^{4} – 3x^{2} + 4x + 5, g(x) = x^{2} + 1 –x

Dividend p(x) = x^{4} – 3x^{2} + 4x + 5

Divisor g(x) = x^{2} + 1 –x

Since degree (4) = 0 < 3 = degree (x^{2} + 1 -x)

Therefore, quotient = x^{2} + x -3, remainder = 8

**(iii)** p(x) = x^{4} – 5x + 6, g(x) = 2 – x^{2}

since degree (4) = 0 < 3 = degree (2 – x^{2})

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

**(2)** Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial:

**(i)** t^{2} – 3, 2t^{4} + 3t^{3} – 2t^{2} – 9t – 12

**(ii)** x^{2} + 3x + 1, 3x^{4} + 5x^{3} – 7x^{2} + 2x + 2

**(iii)** x^{3} – 3x + 1, x^{5} – 4x^{3} + x^{2} + 3x + 1

**Ans**–

**(i)** t^{2} – 3, 2t^{4} + 3t^{3} – 2t^{2} – 9t – 12

Division of the second polynomial by the first polynomial is showing below-

Remainder is zero.

Hence, the first polynomial (t^{2} – 3) is a factor of the second polynomial.

**(ii)** x^{2} + 3x + 1, 3x^{4} + 5x^{3} – 7x^{2} + 2x + 2

Division of the second polynomial by the first polynomial is showing below-

Remainder is zero.

Hence, the first polynomial (x^{2} + 3x + 1) is a factor of the second polynomial 3x^{4} + 5x^{3} – 7x^{2} + 2x + 2.

**(iii)** x^{3} – 3x + 1, x^{5} – 4x^{3} + x^{2} + 3x + 1

Division of the second polynomial by the first polynomial is showing below-

Remainder is not zero,ie 2.

Hence, the first polynomial (x^{3} – 3x + 1) is not a factor of the second polynomial x^{5} – 4x^{3} + x^{2} + 3x + 1.

**(4)** On dividing x^{3} – 3x^{2} + x + 2 by a polynomial g(x), the quotient and remainder were x – 2 and -2x + 4, respectively. Find g(x).

**Ans- **

Dividend = x^{3} – 3x^{2} + x + 2

Divisor = g(x)

Quotient = x – 2

Remainder = -2x + 4

Dividend = Divisor Quotient + Remainder

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} + x + 2 + 2x – 4 = g(x) × (x – 2)

x^{3} – 3x^{2} + 3x -2 = g(x) × (x – 2)

Hence, g(x) is x^{2} – x + 1.

**(5)** Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm and

**(i)** deg p(x) = deg q(x) **(ii)** deg q(x) = deg r(x) **(iii) **deg r(x) =0

**Ans-**

**(i)** deg p(x) = deg q(x)

Let 2x^{2} + 6x – 8 is divided by 2

p(x) = 2x^{2} + 6x – 8

g(x) = 2

q(x) = x^{2} + 3x – 4

r(x) = 0

deg p(x) = deg q(x) =2

Division algorithm

p(x) = g(x) q(x) + r(x)

2x^{2} + 6x – 8 = 2 (x^{2} + 3x – 4) + 0

2x^{2} + 6x – 8 = 2x^{2} + 6x – 8

LHS = RHS

Hence, division algorithm verified.

**(ii)** deg q(x) = deg r(x)

Let x^{3} – 3x^{2} + 5x -3 is divided by x^{2} – 2

p(x) = x^{3} – 3x^{2} + 5x -3

g(x) = x^{2} – 2

q(x) = x – 3

r(x) = 7x – 9

deg q(x) = deg r(x) =1

Division algorithm

p(x) = g(x) q(x) + r(x)

x^{3} – 3x^{2} + 5x -3 = (x^{2} – 2) (x – 3) + 7x – 9

x^{3} – 3x^{2} + 5x -3 = x^{3} – 3x^{2} – 2x + 6 + 7x – 9

x^{3} – 3x^{2} + 5x -3 = x^{3} – 3x^{2} + 5x -3

LHS = RHS

Hence, division algorithm verified.

**(iii)** deg r(x) =0

Let x^{2} + x + 2 is divided by x

p(x) = x^{2} + x + 2

g(x) = x

q(x) = x + 1

r(x) = 2

Therefore, deg r(x) =0

Division algorithm

p(x) = g(x) q(x) + r(x)

x^{2} + x + 2 = (x) (x + 1) + 2

x^{2} + x + 2 = x^{2} + x + 2

LHS = RHS

Hence, division algorithm verified.

**Exercise 2.4**

**(1)** Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case:

**(ii)** x^{3} – 4x^{2} + 5x – 2; 2, 1, 1

The value of zeroes of polynomial x^{3} – 4x^{2} + 5x – 2 is

x – 2 = 0, ie, x = 2

x – 1 + 0, ie, x = 1

and

x – 1 = 1

Put value of zeroes in given polynomial

p(x) = x^{3} – 4x^{2} + 5x – 2

p(2) = 2^{3} – 4(2)^{2} + 5(2) – 2

= 8 – 16 + 10 – 2

= 0

p(1) = x^{3} – 4x^{2} + 5x – 2

p(1) = (1)^{3} – 4(1)^{2} + 5(1) – 2

= 1 – 4 + 5 – 2 = 0

Hence, numbers are zeroes of the given polynomial.

Now, compare the given polynomial with ax^{3} + bx^{2} + cx + d, we get

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

The zeroes of polynomial x^{3} – 4x^{2} + 5x – 2 are 2 , 1 and 1.

Hence, relationship between the zeroes and the coefficients is verified.

**(2)** Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, -7, -14 respectively.

**Ans-**

Given:

α + β + γ = 2

αβ + βγ + γα = -7

αβγ = -14

But, we know

**(3)** If the zeroes of the polynomial x^{3} – 3x^{2} + x +1 are a – b, a, a + b, find a and b.

**Ans**–

Comparing given polynomial to a_{1} x^{3} + b_{1} x^{2} + c_{1} x + d_{1}

a_{1} = 1, b_{1} = -3 c_{1} = 1 and d_{1} = 1

So, x^{4} – 6x^{3} – 26x^{2} + 138x – 35 =(x^{2} – 4x + 1)(x^{2} – 2x – 35)

And

x^{2} – 2x – 35

= x^{2} – 7x + 5x – 35

= x(x -7) +3(x – 7)

=(x – 7) (x + 3)

Therefore, zeroes are x = 7 and x = – 3

Hence, zeroes of the given polynomials are 2 +√3 , 2 -√3 , 7 and – 3.

**(5)** If the polynomial x^{4} – 6x^{3} + 16x^{2} – 25x + 10 is divided by another polynomial x^{2} – 2x + k, the remainder comes out to be x + a, find k and a.

**Ans-**

Divide polynomial x^{4} – 6x^{3} + 16x^{2} – 25x + 10 by another polynomial x^{2} – 2x + k

Comparing coefficient’s of it (2k – 9)x-(8-k)k + 10

with x + a

2k – 9 = 1

K = 5

And

-(8 – k)k + 10= a

Put value of k

-(8-5)5 + 10 = a

a = -15 + 10

= -5

Hence, value of k = 5 and value of a = -5.

**Helping Topics**