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 the part of NCERT Solutions for Class
9 Maths. Here you can find the NCERT Solutions for Class 9 Maths Chapter 2
Polynomials Ex 2.3.
- NCERT Solutions for Class 9 Maths Chapter 2 Number Systems Ex 2.1
- NCERT Solutions for Class 9 Maths Chapter 2 Number Systems Ex 2.2
- NCERT Solutions for Class 9 Maths Chapter 2 Number Systems Ex 2.3
- NCERT Solutions for Class 9 Maths Chapter 2 Number Systems Ex 2.4
- NCERT Solutions for Class 9 Maths Chapter 2 Number Systems Ex 2.5
Ex
2.3 Class 9 Maths Question 1.
Find the remainder when x3 +
3x2 + 3x + 1 is divided by
(i)
x + 1
(ii)
x – ½
(iii)
x
(iv)
x + π
(v)
5 + 2x
Solution:
Let p(x) = x3 + 3x2 +
3x + 1
(i)
The zero of x + 1 is -1.
∴ p(-1) = (-1)3 + 3(-1)2
+ 3(-1) + 1
= -1 + 3 – 3 + 1 = 0
Thus, the required remainder = 0
(ii) The zero of x – ½ is ½.
Thus, the required remainder = 27/8
(iii)
The zero of x is 0.
∴ p(0) = (0)3 + 3(0)2 + 3(0) + 1
= 0 + 0 + 0 + 1 = 1
Thus, the required remainder = 1.
(iv)
The zero of x + π is -π.
p(-Ï€) = (-Ï€)3 + 3(-Ï€)2 + 3(-Ï€) + 1
= -Ï€3 + 3Ï€2 +
(-3Ï€) + 1
= -Ï€3 + 3Ï€2 –
3Ï€ + 1
Thus, the required remainder is -Ï€3 + 3Ï€2 – 3Ï€ +
1.
(v) The zero of 5 + 2x is −5/2.
Thus, the required remainder is −27/8.
Ex 2.3 Class 9 Maths Question 2.
Find the remainder when x3 – ax2 +
6x – a is divided by x – a.
Solution:
We have, p(x) = x3 – ax2 +
6x – a and zero of x – a is a.
∴ p(a) = (a)3 – a(a)2 + 6(a) – a
= a3 – a3 +
6a – a = 5a
Thus, the required remainder is 5a.
Ex
2.3 Class 9 Maths Question 3.
Check
whether 7 + 3x is a factor of 3x3 + 7x.
Solution:
We have, p(x) = 3x3 + 7x
and zero of 7 + 3x is −7/3.
Since, (−490/9) ≠ 0, i.e.
the remainder is not 0.
∴ 3x3 + 7x is not
divisible by 7 + 3x.
Thus, 7 + 3x is not a factor of 3x3 +
7x.
NCERT Solutions for Maths Class 10