NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.3

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) = x³ – 3x² + 5x -3, g(x) = x²-2
(ii) p(x) =x⁴ – 3x² + 4x + 5, g(x) = x² + 1 -x
(iii) p(x) = x⁴ – 5x + 6, g(x) = 2 -x²
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) = x⁴– 3x² + 4x + 5 and g(x) = x² – 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) = x² + x – 3 and g(x) = – x² + 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) t² – 3, 2t⁴ + t³ – 2t² – 9t – 12
(ii) x² + 3x + 1, 3x⁴ + 5x³ -7x² + 2x + 2
(iii) x³ – 3x + 1, x⁵ – 4x³ + x² + 3x + l
Solution:
We have,
P(t) = 2t⁴ + 3t³ – 2t² – 9t – 12
q(x) = t² – 3
By actual division, we have

Here, remainder is zero.
Therefore, q(x) = t² – 3 is the factor of p(x) = 2t⁴ + 3t³ – 2t² – 9t – 12.

(ii) We have,
p(x) = 3x⁴ + 5x³ – 7x² + 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) = 3x⁴ + 5x³ – 7x² + 2x + 2.

(iii) We have,
p(x) = x⁵ – x³ + x² + 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) = x⁵ – x³ + x² + 3x + 1.

Question 3.
Obtain all other zeroes of 3x⁴ + 6x³ – 2x² – 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) = 3x⁴ + 6x³ – 2x² – 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, 3x⁴ + 6x³ – 2x² – 10x – 5

If 3x⁴ + 6x³ – 2x² – 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) = 3x⁴ + 6x³ – 2x² – 10x – 5 are – \(\sqrt { \frac { 5 }{ 3 } }\), \(\sqrt { \frac { 5 }{ 3 } }\), -1 and -1.

Question 4.
On dividing x³ – 3x² + 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 x³– 3x² + x + 2 = g(x) (x – 2) + (-2x + 4)
or x³ – 3x² + x + 2 + 2 x- 4 = g(x) x (x-2)
or x³ – 3x² + 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) = 2x²– 2x + 14; g(x) = 2
q(x) = x² – x + 7 r(x) = 0
Here, deg p(x) = deg q(x)

(ii) deg q(x) = deg r(x)
Polynomial p(x) = x³+ x² + x + 1; g(x) = x² – 1,
q(x) = x + 1, r(x) = 2x + 2

(iii) deg r(x) is 0.
Polynomial p(x) = x²+ 2x² – x + 2; g(x) = x² – 1, q(x) = x + 1, r(x) = 4