NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.3

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.3 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.3.

• Polynomials Class 9 Ex 2.1
• Polynomials Class 9 Ex 2.2
• Polynomials Class 9 Ex 2.3
• Polynomials Class 9 Ex 2.4
• Polynomials Class 9 Ex 2.5

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.3

Ex 2.3 Class 9 Maths Question 1.
Find the remainder when x3 + 3x2 + 3x + 1 is divided by
(i) x + 1
(ii) x- $\frac { 1 }{ 2 }$
(iii) x
(iv) x+π
(v) 5+2x
Solution:
(1) By remainder theorem, the required remainder is equal to p(-1).
Now, p(x)= x3+3x2+3x+1
p(-1) = (-1)3 + 3(-1)2 + 3(-1) +1 =-1+3-3+1=0
Hence, the required remainder = p(-1) = 0

Ex 2.3 Class 9 Maths Question 2.
Find the remainder when x3 – ax2 + 6x – a is divided by x – a.
Solution:
Let p(x) = x3-ax2+6x-a
By remainder theorem, when p(x) is divided by x – a,
Then,remainder = p(a)
p(a)= a3-a . a2+6a-a
=a3 – a3 +6a-a = 5a

Ex 2.3 Class 9 Maths Question 3.
Check, whether 7 + 3x is a factor of 3x3 + 7x.
Solution:
Let f(x) = 3x3 + 7x and g(x) = 7 + 3x
On putting g(x) = 0, we get
7 + 3x = 0 3x = – 7
=> x = $\frac { -7 }{ 3 }$
Thus, zero of g(x) is x =$\frac { -7 }{ 3 }$.

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