# NCERT Solutions Class 9 Maths Chapter 2 Polynomials Exercise 2.3

## Introduction:

In this exercise/article we will learn about Polynomials. Remainder theorem, is it not a simple way to find the remainder obtained on dividing a polynomial by a linear polynomial? We will now generalise this fact in the form of the following theorem. We will also show you why the theorem is true, by giving you a proof of the theorem.

NCERT Class 9 Maths Chapter 2 Polynomials :

- NCERT Class 9 Maths Chapter 2 Polynomials Exercise 2.1
- NCERT Class 9 Maths Chapter 2 Polynomials Exercise 2.2
- NCERT Class 9 Maths Chapter 2 Polynomials Exercise 2.3
- NCERT Class 9 Maths Chapter 2 Polynomials Exercise 2.4
- NCERT Class 9 Maths Chapter 2 Polynomials Exercise 2.5
- Extra Questions Class 9 Maths Chapter 2 Polynomials

**Class 9 Maths Exercise 2.3 (Page-40)**

**Q1. **Find the remainder when x^{3} + 3x^{2} + 3x + 1 is divided by

**(i) **x + 1

**Solution :**

Let p(x) = x^{3} + 3x^{2} + 3x + 1

So, x + 1 = 0

x = - 1

Using remainder theorem,

when p(x) = x^{3} + 3x^{2} + 3x + 1

than p ( - 1 ) = ( - 1 )^{3} + 3( - 1 )^{2} + 3 × - 1 + 1

= - 1 + 3 × 1 - 3 + 1

= - 1 + 3 - 3 + 1 [ because - 1 + 1 = 0 and 3 - 3 = 0 ]

= 0

So, remainder = 0.

**(ii) **x - \(\displaystyle \frac{1}{2}\)

**Solution :**

Let p(x) = x^{3} + 3x^{2} + 3x + 1

So, x - \(\displaystyle \frac{1}{2}\) = 0

x = \(\displaystyle \frac{1}{2}\)

Using remainder theorem,

When p(x) = x^{3} + 3x^{2} + 3x + 1

than p( \(\displaystyle \frac{1}{2}\) ) = ( \(\displaystyle \frac{1}{2}\) )^{3} + 3( \(\displaystyle \frac{1}{2}\) )^{2} + 3 × \(\displaystyle \frac{1}{2}\) + 1

= \(\displaystyle \frac{1}{8}\) + 3 × \(\displaystyle \frac{1}{4}\) + \(\displaystyle \frac{3}{2}\) + 1

= \(\displaystyle \frac{1}{8}\) + \(\displaystyle \frac{3}{4}\) + \(\displaystyle \frac{3}{2}\) + 1

= \(\displaystyle \frac{{1\,+\,6\,+\,12\,+\,8}}{8}\)

= \(\displaystyle \frac{{27}}{8}\)

So, remainder = \(\displaystyle \frac{{27}}{8}\) .

**(iii) **x

**Solution :**

Let p(x) = x^{3} + 3x^{2} + 3x + 1

So, x = 0

x = 0

Using remainder theorem,

when p(x) = x^{3} + 3x^{2} + 3x + 1

than p( 0 ) = ( 0 )^{3} + 3( 0 )^{2} + 3 × 0 + 1

= ( 0 )^{3} + 3 × ( 0 )^{2} + 0 + 1

= 0 + 3 × 0 + 1

= 0 × 0 + 1

= 1

So, remainder = 1 .

**(iv)** x + \(\displaystyle \pi \)

**Solution :**

Let p(x) = x^{3} + 3x^{2} + 3x + 1

So, x + \(\displaystyle \pi \) = 0

x = - \(\displaystyle \pi \)

Using remainder theorem,

when p(x) = x^{3} + 3x^{2} + 3x + 1

than p( - \(\displaystyle \pi \) ) = ( - \(\displaystyle \pi \) )^{3} + 3( - \(\displaystyle \pi \) )^{2} + 3 × - \(\displaystyle \pi \) + 1

= - \(\displaystyle \pi \)^{3} + 3 × \(\displaystyle \pi \)^{2} - 3\(\displaystyle \pi \) + 1

= - \(\displaystyle \pi \)^{3} + 3 - \(\displaystyle \pi \)^{2} - 3\(\displaystyle \pi \) + 1

So, remainder = - \(\displaystyle \pi \)^{3} + 3 - \(\displaystyle \pi \)^{2} - 3\(\displaystyle \pi \) + 1

**(v) **5 + 2x

**Solution :**

Let p(x) = x^{3} + 3x^{2} + 3x + 1

So, 5 + 2x = 0

x = \(\displaystyle -\,\frac{5}{2}\)

Using remainder theorem,

when p(x) = x^{3} + 3x^{2} + 3x + 1

than p( \(\displaystyle -\,\frac{5}{2}\) ) = ( \(\displaystyle -\,\frac{5}{2}\) )^{3} + 3(\(\displaystyle -\,\frac{5}{2}\) )^{2} + 3 × \(\displaystyle -\,\frac{5}{2}\) + 1

= \(\displaystyle -\,\frac{{125}}{8}\,+\,3\,\times \,\frac{{25}}{4}\,-\,\frac{{15}}{2}\,+\,1\)

= \(\displaystyle -\,\frac{{125}}{8}\,+\,\frac{{75}}{4}\,-\,\frac{{15}}{2}\,+\,1\)

= \(\displaystyle \frac{{-\,125\,+\,150\,-\,60\,+\,8}}{8}\)

= \(\displaystyle \frac{{-\,185\,+\,158}}{8}\)

= \(\displaystyle -\,\frac{{27}}{8}\)

So, remainder = \(\displaystyle -\,\frac{{27}}{8}\) .

# NCERT Solutions Class 9 Maths

**Q2. **Find the remainder when x^{3} - ax^{2} + 6x - a is divided by x - a .

**Solution :**

Let p(x) = x^{3} - ax^{2} + 6x - a

So, x - a = 0

= x = a

Using remainder theorem,

when p(x) = x^{3} - ax^{2} + 6x - a

than p( a ) = ( a )^{3} - a( a )^{2} + 6 × a - a

= a^{3} - a × a^{2} + 6a - a

= a^{3} - a^{3} + 6a - a [ because a^{3} - a^{3} = 0 ]

= 6a - a

= 5a

So, remainder = 5a .

**Q3. **Check whether 7 + 3x is a factor of 3x^{3} + 7x.

**Solution :**

Let p(x) = 3x^{3} + 7x

So, 7 + 3x = 0

3x = - 7

x = \(\displaystyle -\,\frac{7}{3}\)

Using remainder theorem,

when p(x) = 3x^{3} + 7x

than p( \(\displaystyle -\,\frac{7}{3}\) ) = 3x^{3} + 7x

= 3 ( \(\displaystyle -\,\frac{7}{3}\) )^{3} + 7 × \(\displaystyle -\,\frac{7}{3}\)

= \(\displaystyle 3\,\times \,-\,\frac{{343}}{{27}}\,-\,\frac{{49}}{3}\)

=

= \(\displaystyle -\,\frac{{343}}{9}\,-\,\frac{{49}}{3}\)

= \(\displaystyle \frac{{-\,343\,-\,147}}{9}\)

= \(\displaystyle -\,\frac{{490}}{9}\)

So, 7 + 3x is not a factor of 3x^{3} + 7x .

NCERT Class 9 Maths Chapter 2 Polynomials :

- NCERT Class 9 Maths Chapter 2 Polynomials Exercise 2.1
- NCERT Class 9 Maths Chapter 2 Polynomials Exercise 2.2
- NCERT Class 9 Maths Chapter 2 Polynomials Exercise 2.3
- NCERT Class 9 Maths Chapter 2 Polynomials Exercise 2.4
- NCERT Class 9 Maths Chapter 2 Polynomials Exercise 2.5