NCERT Solutions for Class 9 Maths Chapter 2: Polynomials - Exercise 2.3

1. Find the remainder when \[\mathbf{{{x}^{3}}+3{{x}^{2}}+3x+1}\] is divided by

(i) \[\mathbf{x+1}\] 

(ii) $\mathbf{x-\dfrac{1}{2}}$  

(iii) \[\mathbf{x}\] 

(iv) \[\mathbf{x+\pi }\] 

(v) \[\mathbf{5+2x}\] 

Ans: Remainder is the amount left over after performing a computation of division.

(i) \[\mathbf{{{x}^{3}}+3{{x}^{2}}+3x+1\div x+1}\]

We will use the long division method.

\[\begin{align}&x+1\overset{{{x}^{2}}+2x+1}{\overline{\left){{{x}^{3}}+3{{x}^{2}}+3x+1}\right.}} \\   &\quad\quad\:\:\text{       }{{x}^{3}}+\text{ }{{x}^{2}}\text{ } \\  &\quad\quad\:\text{    }\underline{\text{ }-\text{          }-\text{                   }} \\  &\quad\quad\quad\quad\text{               }2{{x}^{2}}+3x+1 \\  &\quad\quad\quad\quad\text{               }2{{x}^{2}}+2x \\ &\quad\quad\quad\quad\text{           }\underline{\text{ }-\text{           }-\text{           }} \\  &\quad\quad\quad\quad\quad\quad\quad x+1 \\  &\quad\quad\quad\quad\quad\quad\quad x+1 \\  & \quad\quad\quad\quad\quad \quad\quad\underline{-\text{        }-\text{   }} \\  &\quad\quad\quad\quad\quad\quad\quad\quad\text{  }\underline{\text{       }0\text{       }} \\ \end{align}\]

Therefore, the remainder obtained is \[0\] .

(ii) \[\mathbf{\text{ }{{x}^{3}}+3{{x}^{2}}+3x+1\div x-\dfrac{1}{2}}\]

We will use the long division method.

$\begin{align}   & x-\dfrac{1}{2}\overset{{{x}^{2}}+\dfrac{7}{2}x+\dfrac{19}{4}}{\overline{\left){{{x}^{3}}+3{{x}^{2}}+3x+1}\right.}} \\  &\quad\quad\quad\:{{x}^{3}}-\dfrac{{{x}^{2}}}{2} \\  & \quad\quad\quad\underline{-\text{                  + }  \qquad} \\  & \quad\quad\quad\quad\quad\quad\dfrac{7}{2}{{x}^{2}}+3x+1 \\  & \quad\quad\quad\quad\quad\quad\dfrac{7}{2}{{x}^{2}}-\dfrac{7}{4}x \\  & \quad\quad\quad\quad\quad\quad\underline{-\text{            }  + \quad} \\  & \quad\quad\quad\quad\quad\quad\quad\quad\dfrac{19}{4}x+1 \\  & \quad\quad\quad\quad\quad\quad\quad\quad\dfrac{19}{4}x-\dfrac{19}{8} \\  & \quad\quad\quad\quad\quad\quad\quad\quad\underline{\text{ }-\text{        }+\text{       }} \\  & \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\underline{\text{          }\dfrac{27}{8}\text{     }} \\ \end{align}$  

Therefore, the remainder obtained is $\dfrac{27}{8}$ .

(iii) \[\mathbf{{{x}^{3}}+3{{x}^{2}}+3x+1\div x}\] 

We will use the long division method.

$\begin{align}   & x\overset{{{x}^{2}}+3x+3}{\overline{\left){{{x}^{3}}+3{{x}^{2}}+3x+1}\right.}} \\  & \quad {{x}^{3}} \\  & \underline{-\text{                            }\qquad \qquad}\text{ } \\  & \quad\quad\quad 3{{x}^{2}}+3x+1 \\  & \quad\quad\quad 3{{x}^{2}} \\  & \quad\quad\text{      }\underline{\text{ }-\text{                    }\qquad \qquad} \\  & \quad\quad\quad\quad\quad\quad 3x+1 \\  & \quad\quad\quad\quad\quad\quad 3x \\  & \quad\quad\quad\quad\quad\quad\text{                }\underline{-\text{             }\qquad} \\  & \quad\quad\quad\quad\quad\quad\quad\quad\underline{\text{      }1\text{        }} \\ \end{align}$

Therefore, the remainder obtained is $1$ .

(iv) \[\mathbf{{{x}^{3}}+3{{x}^{2}}+3x+1\div x+\pi} \]

We will use the long division method.

$\begin{align}   & x+\pi \overset{{{x}^{2}}+(3-\pi )x+(3-3\pi +{{\pi }^{2}})}{\overline{\left){{{x}^{3}}+3{{x}^{2}}+3x+1}\right.}} \\  & \quad\quad\quad {{x}^{3}}+\pi {{x}^{2}} \\  & \quad\quad\quad\underline{-\text{    }-\text{                                              }\qquad \qquad} \\  & \quad\quad\quad\quad\quad (3-\pi ){{x}^{2}}+3x+1 \\  & \quad\quad\quad\quad\quad (3-\pi ){{x}^{2}}+(3-\pi )\pi x \\ & \quad\quad\quad\quad\quad\text{                         }\underline{-\quad \qquad \quad\text{               }-\text{                                      }\qquad \qquad} \\  & \quad\quad\quad\quad\quad\quad\quad\left[ 3-3\pi +{{\pi }^{2}} \right]x+1 \\  & \quad\quad\quad\quad\quad\quad\quad\left[ 3-3\pi +{{\pi }^{2}} \right]x+\left[ 3-3\pi +{{\pi }^{2}} \right]\pi  \\  & \quad\quad\quad\quad\quad\quad\quad\text{                     }\underline{-\quad\quad\qquad\qquad\text{                           }-\text{                               }\qquad \qquad\quad}\text{ } \\  &\quad\quad\quad\quad\quad\quad\quad\quad\quad\underline{\text{                  }\left[ 1-3\pi +3{{\pi }^{2}}-{{\pi }^{3}} \right]\text{               }} \\ \end{align}$

Therefore, the remainder obtained is $-{{\pi }^{3}}+3{{\pi }^{2}}-3\pi +1$  .

(v) \[\mathbf{{{x}^{3}}+3{{x}^{2}}+3x+1\div ~5+2x}\] 

We will use the long division method.

$\begin{align}   & 2x+5\overset{\dfrac{{{x}^{2}}}{2}+\dfrac{x}{4}+\dfrac{7}{8}}{\overline{\left){{{x}^{3}}+3{{x}^{2}}+3x+1}\right.}} \\  & \quad\quad\quad{{x}^{3}}+\dfrac{5}{2}{{x}^{2}} \\  & \quad\quad\quad\text{      }\underline{-\text{       }\,-\text{                      }\quad\quad\qquad\qquad} \\  & \quad\quad\quad\quad\quad \dfrac{{{x}^{2}}}{2}+3x+1 \\  & \quad\quad\quad\quad\quad \dfrac{{{x}^{2}}}{2}+\dfrac{5x}{4} \\  & \quad\quad\quad\quad\quad\text{            }\underline{-\text{       }\,-\text{                }\quad \quad \qquad\qquad} \\  & \quad\quad\quad\quad\quad\quad\quad\dfrac{7x}{4}+1 \\  &\quad\quad\quad\quad\quad\quad\quad\dfrac{7x}{4}+\dfrac{35}{8} \\  & \quad\quad\quad\quad\quad\quad\quad\text{                    }\underline{-\text{       }\,\,\,\,-\text{         }\quad\quad\qquad\qquad} \\  & \quad\quad\quad\quad\quad\quad\quad\quad\quad\underline{\text{        }-\dfrac{27}{8}\text{   }} \\ \end{align}$ 

Therefore, the remainder obtained is $-\dfrac{27}{8}$  .

2. Find the remainder when  \[\mathbf{{{x}^{3}}-a{{x}^{2}}+6x-a}\] is divided by \[\mathbf{x-a}\].

Ans: Remainder is the amount left over after performing a computation of division.

\[{{x}^{3}}-a{{x}^{2}}+6x-a\div x-a\]

Here, we will use a long division method to find the remainder.

$\begin{align}   & x-a\overset{{{x}^{2}}+6}{\overline{\left){{{x}^{3}}-a{{x}^{2}}+6x-a}\right.}} \\  & \quad\quad\quad {{x}^{3}}-a{{x}^{2}} \\  & \quad\quad\text{   }\underline{\text{ }-\text{    }+\text{                      }\qquad \qquad} \\  & \quad\quad\quad\quad\quad \qquad 6x-a \\  & \quad\quad\quad\quad\quad \qquad6x-6a \\  & \quad\quad\quad\quad\qquad\text{                     }\underline{\text{ }-\text{    }+\text{     }\qquad \qquad} \\  & \quad\quad\quad\quad\quad\qquad\qquad\underline{\text{           }5a\text{   }} \\ \end{align}$

Hence, when we divide \[{{x}^{3}}-a{{x}^{2}}+6x-a\] by\[x-a\] , the remainder we obtain is $5a$.

3. Check whether 7+3x is a factor of \[\mathbf{3{{x}^{3}}+7x}\]. 

Ans: A factor is a number that divides into another number exactly without leaving a remainder.

Let us divide \[\left( 3{{x}^{3}}+7x \right)\] by\[(7+3x)\] .

Here, we will use a long division method to find the remainder.

We will write $\left( 3{{x}^{3}}+7x \right)$ as $\left( 3{{x}^{3}}+0{{x}^{2}}+7x \right)$ for easy solving the question.

$\begin{align}  & 3x+7\overset{{{x}^{2}}-\dfrac{7}{3}x+\dfrac{70}{9}}{\overline{\left){3{{x}^{3}}+0{{x}^{2}}+7x}\right.}} \\  & \quad\quad\quad 3{{x}^{3}}+7{{x}^{2}} \\  & \quad\quad\quad\text{        }\underline{-\text{    }\,\,-\text{                    }\qquad\qquad} \\  & \quad\quad\quad\quad\quad-7{{x}^{2}}+7x \\  & \quad\quad\quad\quad\quad-7{{x}^{2}}-\dfrac{49}{3}x \\ & \quad\quad\quad\quad\quad\quad\text{                 }\underline{+\text{     }\,\,+\text{           }\qquad\qquad} \\  & \quad\quad\quad\quad\quad\quad\quad\dfrac{70x}{3} \\  & \quad\quad\quad\quad\quad\quad\quad\dfrac{70x}{3}+\dfrac{490}{9} \\  & \quad\quad\quad\quad\quad\quad\quad\quad\text{                         }\underline{-\text{       }\,\,-\text{            }\qquad\qquad} \\  & \quad\quad\quad\quad\quad\quad\quad\quad\quad\underline{\text{           }-\dfrac{490}{9}\text{    }} \\ \end{align}$ 

The remainder obtained is $-\dfrac{490}{9}$ which is not zero.

Therefore, we can say that \[7+3x\] is not a factor of  \[3{{x}^{3}}+7x\].

Before delving into Polynomials, let us first understand the below terminologies- 

Variables and Constants

Check out the following equation

3X + 5X = 16

3X + 5X = 32

What you see here is that the equation is the same, but the result is different. When is that possible? When the values of X are different in the two cases. In the first equation, the value of X is 2 while in the second, the value is 4. This is called variable - one that varies in different cases.

Now look at 3 and 5 in the above equations. In both cases the values of 3 and 5 remain the same. Multiplying three X will always show you the result when three X are added. Again, multiplying five X will always show you the result of the addition of five X. These are called constants - things that remain constantly remain the same in all the different cases.

Terms

Look at the following equations-

5X - 2X = 6

Here, the equation has two parts- 5X and 2X. The parts of an equation are called terms.

Polynomial

Polynomial is a mathematical expression which consists of terms. These terms are made up of constants, variables and non-negative or non-fractionated exponents. Polynomials only include addition, subtraction, and multiplication.

Examples of Polynomials 

5X +3 

3Y - 7

6Y * 8 or 48Y or 48* Y

X³ + 3X² +X

5X + 3X - 6X

And other such terms

But X + 1∕X is not a polynomial since 1/x can be written as X⁻¹ . As said earlier, a polynomial does not allow negative exponents. Similarly, variable with roots cannot fall under Polynomials since √X can be written as X^½

What You Will Learn In Class 9 Maths Chapter 2 Exercise 2.3

In Polynomial Class 9 Ex 2.3, you will learn how to divide a Polynomial with another Polynomial or with any term. This will help you find the remainders of the division and will also tell you if one Polynomial is a factor of the other.

For example in our NCERT Maths Class 9 Exercise 2.3 Solutions , question number 1 (v) tells us to find the remainder when X³ + 3X² + 3X + 1 is divided by 5 + 2X

We employed Long division method and divided the first Polynomial with the second. We went on dividing the Polynomial until the variable was finally eliminated. Anything that remains as remainder after the variable can be considered as remainder.

Now, if anything remains as the remainder by dividing the first Polynomial by the second, then the second is not the factor of the first. If the remainder is 0, then the second is the factor of the first.

NCERT Solutions Class 9 Maths Chapter 2 Exercises

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NCERT Solutions for Class 9 Maths All Chapters

• Chapter 1 - Number System

• Chapter 2 - Polynomials 

• Chapter 3 - Coordinate Geometry

• Chapter 4 - Linear Equations in Two Variables

• Chapter 5 - Introductions to Euclids Geometry

• Chapter 6 - Lines and Angles

• Chapter 7 - Triangles

• Chapter 8 - Quadrilaterals

• Chapter 9 - Areas of Parallelogram and Triangles

• Chapter 10 - Circles

• Chapter 11 - Constructions

• Chapter 12 - Herons formula

• Chapter 13 - Surface area and Volumes

• Chapter 14 - Statistics

• Chapter 15 - Probability