NCERT Solutions Class 9 Maths Chapter 2 Polynomials Ex 2.3

NCERT Solutions Class 9 Maths 
Chapter – 2 (Polynomials) 

The NCERT Solutions in English Language for Class 9 Mathematics Chapter – 2 Polynomials Exercise 2.3 has been provided here to help the students in solving the questions from this exercise. 

Chapter 2: Polynomials 

• NCERT Solution Class 9 Maths Ex – 2.1
• NCERT Solution Class 9 Maths Ex – 2.2
• NCERT Solution Class 9 Maths Ex – 2.4
• NCERT Solution Class 9 Maths Ex – 2.5

Exercise – 2.3 

1. Find the remainder when x³ + 3x² + 3x + 1 is divided by
(i) x + 1
(ii) x − 1/2
(iii) x
(iv) x + π

Answer –
(i) x + 1
The zero of x + 1 = 0
⇒ x = −1
Let p(x) = x³ + 3x² + 3x +1
∴ p(-1) = (-1)3 + 3(-1)2 + 3(-1) +1
⇒ -1 + 3- 3 + 1
⇒ 0
Thus, the required remainder = 0

(ii) x−1/2
The zero of x − 1/2 = 0
⇒ x = 1/2
Let p(x) = x³ + 3x² + 3x +1
p(1/2) = (1/2)³ + 3(1/2)² + 3(1/2) + 1
⇒ (1/8) + (3/4) + (3/2) + 1
⇒                             LCM of 8, 4 and 2 = 8

⇒

⇒
Thus, the required remainder =

(iii) x
The zero of x = 0
⇒ x = 0
Let p(x) = x³ + 3x² + 3x +1
p(0) = (0)³+3(0)²+3(0)+1
⇒ 1
Thus, the required remainder = 1.

(iv) x + π
The zero of x + π = 0
⇒ x = – π
Let p(x) = x³ + 3x² + 3x +1
p(π) = (−π)³ +3(−π)² + 3(−π) + 1
⇒ −π³ + 3π² − 3π + 1
Thus, the required remainder is -π³ + 3π² – 3π + 1.

(v) 5 + 2x
The zero of 5 + 2x = 0
⇒ x = – 5/2
Let p(x) = x³ + 3x² + 3x +1
p(-5/2) = (-5/2)³ + 3(-5/2)² + 3(-5/2) + 1
⇒ (-125/8) + (75/4) – (15/2) + 1
⇒                                LCM of 8, 4 and 2 = 8

⇒  

⇒
Thus, the required remainder is .

2. Find the remainder when x³ − ax² + 6x − a is divided by x – a.
Answer –
Let p(x) = x³ − ax² + 6x − a
x − a = 0
∴ x = a
The remainder:
p(a) = (a)³ − a(a²) + 6(a) − a
= a³ − a³ + 6a − a
= 5a

3. Check whether 7 + 3x is a factor of 3x³ + 7x.
Answer –
Let p(x) = 3x³ + 7x
7 + 3x = 0
⇒ 3x = −7
⇒ x = -7/3
∴ The remainder:
p(-7/3) = 3(-7/3)³ + 7(-7/3)
⇒ -(343/9) + (-49/3)

⇒                 LCM of 9 and 3 = 9

⇒

⇒ ≠ 0
∴ 7 + 3x is not a factor of 3x³ + 7x.