NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.3

These NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.3 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Exercise 2.3

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 + p
(v) 5 + 2x
Solution:
(i) By long division, we have:

Here, the remainder is 0.

(ii) By long division, we have:

Here, the remainder is \(\frac{27}{8}\)

(iii) By long division, we have

Here, the remainder is 1

(iv) we have given
x³ + 3x² + 3x + 1 ÷ x + π
or, x³ + 3x² + 3x + 1 ÷ x + \(\frac{22}{7}\) (∵ π = \(\frac{22}{7}\))
By long division,

Here, the remainder is \(-\frac{3669}{49}\)

(v) We have given
x³ + 3x² + 3x + 1 ÷ 5 + 2x
or, x³ + 3x² + 3x + 1 ÷ 2x + 5
By long division, we have

Here, remainder is \(-\frac{27}{8}\)

Question 2.
Find the remainder when x³ – ax² + 6x – a is divided by x – a.
Solution:
Here, p(x) = x³ – ax² + 6x – a, and the zero of x – a is a
So, p(a) = a³ – a(a)² + 6(a) – a
= a³ – a³ + 6a – a
= 5a
So, 5a is the remainder when x³ – ax + 6x – a is divided by x – a.

Question 3.
Check whether 7 + 3x is a factor of 3x³ + 7x.
Solution:
We know that 7 + 3x is a factor of 3x³ + 7x only if 7 + 3x divides 3x³ + 7x leaving no remainder.
Here, p(x) = 3x³ + 7x and the zero of 7 + 3x is \(-\frac{7}{3}\). so,

Here, the remainder is \(\frac{-490}{9}\).
Therefore 7 + 3x is not the factor of 3x³ + 7x.