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Class 9 NCERT Solutions – Chapter 2 Polynomials – Exercise 2.3
  • Last Updated : 23 Feb, 2021

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

(i) x + 1

Solution:

x + 1 = 0

x  = −1

Therefore remainder will be f(x):



f(−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

Therefore remainder will be f(x):

f(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

Therefore remainder will be f(x):

f(0) = (0)3 + 3(0)2 + 3(0) + 1

= 1

(iv) x + pi



Solution:

x + pi = 0

x = −pi

Therefore remainder will be f(x):

f(−pi) = (−pi)3 + 3(−pi)2 + 3(−pi) + 1

= −pi3 + 3pi2 − 3pi + 1

(v) 5 + 2x

Solution:

5 + 2x = 0

2x = −5

 x = -5/2

Therefore remainder will be f(x) :

f(-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:

Let f(x) = x3 − ax2 + 6x − a

x − a = 0

∴ x = a

Therefore remainder will be f(x):

f(a) = (a)3 − a(a2) + 6(a) − a

= a3 − a3 + 6a − a 

= 5a

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

Solution:

7 + 3x = 0

3x = −7

 x = -7/3

Therefore remainder will be f(x):

f(-7/3) = 3(-7/3)3 + 7(-7/3) 

= – (343/9) + (-49/3)

= (-343- (49) * 3)/9

= (-343 – 147)/9

= – 490/9 ≠ 0

∴ 7 + 3x is not a factor of 3x3 + 7x

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