NCERT Solutions Class 9th Maths Chapter – 2 Polynomials Exercise – 2.3

Question 1. Find the remainder when x³ + 3x² + 3x + 1 is divided by

(i) x + 1

Solution:
x + 1 = 0
⇒ x = −1
p(−1) = (−1)³ + 3(−1)² + 3(−1) + 1
⇒ −1 + 3 − 3 + 1
⇒ 0

(ii) x − 1/2

Solution:
x – 1/2 = 0
⇒ x = 1/2
p(1/2) = (1/2)³ + 3(1/2)² + 3(1/2) + 1
⇒ (1/8) + (3/4) + (3/2) + 1
⇒ 27/8

(iii) x

Solution:
x = 0
p(0) = (0)³ + 3(0)² + 3(0) + 1
= 1

(iv) x + π

Solution:
x + π = 0
⇒ x = -π
p(-π) = (−π)³ + 3(−π)² + 3(-π) + 1
= -π³ + 3π² – 3π + 1

(v) 5 + 2x

Solution:
5 + 2x = 0
⇒ 2x = -5
⇒ x = -5/2
p(-5/2) = (-5/2)³ + 3(-5/2)² + 3(-5/2) + 1
= (-125/8) + (75/4) – (15/2) + 1
= -27/8

Question 2. Find the remainder when x³ – ax² + 6x – a is divided by x – a.

Solution:
We have, p(x) = x³ – ax² + 6x – a
x – a = 0
⇒ x = a
p(a) = (a)³ – a(a)² + 6(a) – a
= a³ – a³ + 6a – a
= 5a
Thus, the required remainder is 5a.

Question 3.Check whether 7 + 3x is a factor of 3x³+7x.

Solution:
7+3x = 0
⇒ 3x = −7
⇒ x = -7/3
p(-7/3) = 3(-7/3)³ + 7(-7/3)
= -(343/9)+(-49/3)
= (-343-(49)3)/9
= (-343-147)/9
= -490/9 ≠ 0
Thus, 7 + 3x is not a factor of 3x³ + 7x