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

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

NCERT TEXTBOOK EXERCISES

Ex 2.3 Class 10 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) Here p(x) = x3 -3x2 + 5x – 3 and g(x) = x2 -2
dividing p(x) by g(x) ⇒

Quotient = x – 3, Remainder = 7x – 9

(ii) Here p(x) = x4– 3x2 + 4x + 5 and g(x) = x2 + 1 -x
dividing p(x) by g(x) ⇒

Quotient = x2 + x – 3, Remainder = 8

(iii) Herep(x) = x4– 5x + 6 and g(x) = 2-x2
Rearranging g(x) = -x2 + 2
dividing p(x) by g(x) ⇒
Quotient = -x2 – 2
Remainder = -5x + 10.

Ex 2.3 Class 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:

(i) First polynomial = t2 – 3,
Second polynomial = 2t4 + 3t3 – 2t2 – 9t – 12
dividing second polynomial
by first polynomial ⇒
∵ Remainder is zero.
∴First polynomial is a factor of second polynomial.

(ii) First polynomial = x2 + 3x + 1
Second polynomial = 3x4 + 5x3 – 7x2 + 2x + 2
dividing second polynomial
by first polynomial ⇒
∵ Remainder is zero.
∴ First polynomial is a factor of second polynomial.

(iii) First polynomial = x3 – 3x + 1
Second polynomial = x5 – 4x3 + x2 + 3x + 1.
∵ Remainder ≠ 0.
∴ First polynomial is not a factor of second polynomial.

Ex 2.3 Solutions Class 10 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:

NCERTSolutions Ex 2.3 Class 10 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:

p(x) = x3 – 3 x 2 + x + 2 g(x) = ?
Quotient = x – 2; Remainder = -2x + 4
On dividing p(x) by g(x), we have
p(x) = g(x) x quotient + remainder
⇒ x3– 3x2 + x + 2 = g(x) (x – 2) + (-2x + 4)
⇒ x– 3x2 + x + 2 + 2 x- 4 = g(x) x (x-2)
⇒ x3 – 3x2 + 3x – 2 = g(x) (x – 2)

Exercise 2.3 Class 10 Maths 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) p(x),g(x),q(x),r(x)
deg p(x) = deg q(x)
∴ both g(x) and r(x) are constant terms.
p(x) = 2x2– 2x + 14; g(x) = 2
q(x) = x2 – x + 7; r(x) = 0

(ii) deg q(x) = deg r(x)
∴ this is possible when
deg of both q(x) and r(x) should be less than p(x) and g(x).
p(x) = x3+ x2 + x + 1; g(x) = x2 – 1
q(x) = x + 1, r(x) = 2c + 2

(iii) deg r(x) is 0.
This is possible when product of q(x) and g(c) form a polynomial whose degree is equal to degree of p(x) and constant term.

## NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Exercise 2.3 PDF

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