# Class 9 NCERT Solutions – 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  = âˆ’1

Therefore remainder will be f(x):

f(âˆ’1) = (âˆ’1)3 + 3(âˆ’1)2 + 3(âˆ’1) + 1

= âˆ’1 + 3 âˆ’ 3 + 1

= 0

(ii) x â€“ 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 = âˆ’pi

Therefore remainder will be f(x):

f(âˆ’pi) = (âˆ’pi)3 + 3(âˆ’pi)2 + 3(âˆ’pi) + 1

= âˆ’pi3 + 3pi2 âˆ’ 3pi + 1

(v) 5 + 2x

Solution:

5 + 2x = 0

2x = âˆ’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 âˆ’ ax2 + 6x âˆ’ a is divided by x – a.

Solution:

Let f(x) = x3 âˆ’ ax2 + 6x âˆ’ a

x âˆ’ a = 0

âˆ´ x = a

Therefore remainder will be f(x):

f(a) = (a)3 âˆ’ a(a2) + 6(a) âˆ’ a

= a3 âˆ’ a3 + 6a âˆ’ a

= 5a

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

Solution:

7 + 3x = 0

3x = âˆ’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 â‰  0

âˆ´ 7 + 3x is not a factor of 3x3 + 7x

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