NCERT Solutions For Class 9 Math Chapter – 2 Exercise – 2.3

NCERT Solutions For Class 9 Math Chapter – 2 Exercise – 2.3

 

Q1. Find the remainder  when x3 + 3x2 + 3x + 1 is divided by

  • X + 1

Solution:-

 

X + 1 = 0

X = -1

 

Remainder

P(-1) = (-1)3 + 3(-1)2 + 3(-1) + 1

= -1 + 3 – 3 + 1

= 0

 

  • X – 1/2

Solution:-

 

X – 1/2 = 0

X = 1/2

 

Remainder

P(1/2) = (1/2)3 + 3(1/2)2 + 3(1/2) + 1

= (1/8) + (3/4) + (3/2) + 1

= 27/8

 

  • X

Solution:-

 

X = 0

Remainder:

P(0) = (0)3 + 3(0) + 1

= 1

 

  • X +

Solution:-

 

X + ∏ = 0

X = -∏

 

Remainder

 p(0) = (-∏)3 + 3(-∏)2 + 3(-∏) + 1

= -∏3 + 3∏ + 1

 

  • 5 + 2x

Solution:-

 

5 + 2x = 0

X = -5/2

Remainder

 

(-5/2)3 + 3(-5/2)2 + 3(-5/2) + 1

= -27/8

 

Q2. Find the remainder when x3 – ax2 + 6x – a is divided by x – a.

Solution:-

 

X – a = 0

X = a

 

Remainder

P(a) = (a)3 – a(a2) + 6(a) – a

= a3 – a3 + 6a – a

= 5a

 

Q3. Check  whether 7 + 3x is a factor of 3x3 + 7x.

Solution:-

 

7 + 3x = 0

3x = – 7

X = -7/3

Remainder:

3(-7/3)3 + 7(-7/3) =-(343/9)  + (-49/3)

= (-343 – (49)3)/9

= -490/9 ≠ 0