These NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.3 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 10 Maths Chapter 2 Polynomials

## Exercise 2.3

Question 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}

Solution:

(i)

Therefore,

quotient = x – 3 and remainder = 7x – 9

(ii) First we arrange the terms of the dividend and the divisor in the decreasing order of their degrees.

∴ p(x) = x^{4}– 3x^{2} + 4x + 5 and g(x) = x^{2} – x + 1

Therefore,

quotient = x2 +x-3 and remainder = 8

(iii) First we arrange the terms of the dividend and the divisior in the decreasing order of their degrees.

∴ p(x) = x^{2} + x – 3 and g(x) = – x^{2} + 2

Therefore,

quotient = – x² – 2 and remainder = – 5x + 10

Question 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} + t^{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 + l

Solution:

We have,

P(t) = 2t^{4} + 3t^{3} – 2t^{2} – 9t – 12

q(x) = t^{2} – 3

By actual division, we have

Here, remainder is zero.

Therefore, q(x) = t^{2} – 3 is the factor of p(x) = 2t^{4} + 3t^{3} – 2t^{2} – 9t – 12.

(ii) We have,

p(x) = 3x^{4} + 5x^{3} – 7x^{2} + 2x + 2

and q(x) = x² + 3x + 1

By actual division, we have

Here, remainder is not zero.

Therefore, q(x) = x² – 3x + 1 is not a factor of p(x) = 3x^{4} + 5x^{3} – 7x^{2} + 2x + 2.

(iii) We have,

p(x) = x^{5} – x^{3} + x^{2} + 3x + 1

and q(x) = x³ – 3x + 1

By actual division, we have

Here, remainder is not zero.

Therefore, q(x) = x² – 3x + 1 is not a factor of p(x) = x^{5} – x^{3} + x^{2} + 3x + 1.

It is a remainder theorem calculator that calculates the remainder and quotient in the process of division.

Question 3.

Obtain all other zeroes of 3x^{4} + 6x^{3} – 2x^{2} – 10x – 5, if two of its zeroes are \(\sqrt { \frac { 5 }{ 3 } }\) and – \(\sqrt { \frac { 5 }{ 3 } }\)

Solution:

We have given that two zeroes of polynomial p(x) = 3x^{4} + 6x^{3} – 2x^{2} – 10x – 5 are \(\sqrt { \frac { 5 }{ 3 } }\) and – \(\sqrt { \frac { 5 }{ 3 } }\)

∴ (x – \(\sqrt { \frac { 5 }{ 3 } }\))(x + \(\sqrt { \frac { 5 }{ 3 } }\)) = x² – \(\frac { 5 }{ 3 }\) is a factor is given polynomial p(x).

Now apply the division algorithm to the given polynomial and x² – \(\frac { 5 }{ 3 }\)

So, 3x^{4} + 6x^{3} – 2x^{2} – 10x – 5

If 3x^{4} + 6x^{3} – 2x^{2} – 10x – 5 = 0

Then, (x + \(\sqrt { \frac { 5 }{ 3 } }\))(x – \(\sqrt { \frac { 5 }{ 3 } }\))(x + 1)(3x + 3) = 0

∴ x = \(\sqrt { \frac { 5 }{ 3 } }\) or x = \(\sqrt { \frac { 5 }{ 3 } }\) Therefore, the zeroes of polynomial p(x) = 3x^{4} + 6x^{3} – 2x^{2} – 10x – 5 are – \(\sqrt { \frac { 5 }{ 3 } }\), \(\sqrt { \frac { 5 }{ 3 } }\), -1 and -1.

Question 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).

Solution:

We know that

Dividend = Divisor x Quotiet + Remainder

∴ x² – 3x² + x + 2 = g(x) x (x – 2) + (- 2x + 4)

or x^{3}– 3x^{2} + x + 2 = g(x) (x – 2) + (-2x + 4)

or x^{3 }– 3x^{2} + x + 2 + 2 x- 4 = g(x) x (x-2)

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

Question 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

Solution:

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

Polynomial p(x) = 2x^{2}– 2x + 14; g(x) = 2

q(x) = x^{2} – x + 7 r(x) = 0

Here, deg p(x) = deg q(x)

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

Polynomial p(x) = x^{3}+ x^{2} + x + 1; g(x) = x^{2} – 1,

q(x) = x + 1, r(x) = 2x + 2

(iii) deg r(x) is 0.

Polynomial p(x) = x^{2}+ 2x^{2} – x + 2; g(x) = x^{2} – 1, q(x) = x + 1, r(x) = 4