# NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.3

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## 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

We hope you found these NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.3 helped you with finding the solutions easily.

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