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

Last updated date: 14th Jul 2024
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## NCERT Solutions for Class 9 Maths Chapter 2 (Ex 2.3)

NCERT Solutions for Class 9 Maths Chapter 2 Exercise 2.3 are available on Vedantu.com. This chapter deals with Polynomials. In our Exercise 2.3 NCERT Maths Class 9 Solutions we will show you how you can find leverage long division method in case of polynomials to solve problems. Vedantu provides students with a Free PDF download option for all the NCERT Solutions of updated CBSE Textbooks. Subjects like Science, Maths, Engish will become easy to study if you have access to NCERT Solution for Class 9 Science, Maths solutions, and solutions of other subjects that are available on Vedantu only.

 Class: NCERT Solutions for Class 9 Subject: Class 9 Maths Chapter Name: Chapter 2 - Polynomials Exercise: Exercise - 2.3 Content-Type: Text, Videos, Images and PDF Format Academic Year: 2024-25 Medium: English and Hindi Available Materials: Chapter WiseExercise Wise Other Materials Important QuestionsRevision Notes

### About the Chapter 2 Class 9 Maths - Polynomials

Chapter 2 of NCERT Solutions for Class 9 Maths Polynomials is all about the fundamentals of polynomials, such as the different types of polynomials, finding roots, and solving polynomial equations. Polynomials are algebraic expressions with one or more variables. These NCERT solutions for class 9 of maths Chapter 2 also cover the remainder theorem, factor theory of polynomials, algebraic identities, and polynomials of various degrees. NCERT Solutions for Class 9 Maths -  Polynomials in Chapter 2 exemplify the difference between linear, quadratic, and cubic polynomials. The Remainder Theorem and the Factor Theorem - theorems that help identify the factors of a polynomial, are two important theorems mentioned.

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## Access NCERT Solutions for Class 9 Mathematics Chapter 2 – Polynomials

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

 Chapter 2 - Polynomials All Exercises in PDF Format Exercise 2.1 5 Question & Solutions Exercise 2.2 4 Questions & Solutions Exercise 2.4 5 Questions & Solutions Exercise 2.5 16 Questions & Solutions

### Why Should You Opt For Vedantu

Our NCERT Maths Class 9 Chapter 2 Exercise 2.3 solutions adhere to the NCERT guidelines strictly. The solutions to the problems are explained in detail. As a result, any student who follows our Polynomials Class 9 Exercise 2.3 solutions will find it easy to understand the chapter without having any confusion whatsoever. The answers are not written in a hurried manner. We have given detailed explanations. That is why while writing the answer to Question number 1 (iii), we not only mentioned the result of the division but we showed how to do the long division. We showed how X³ + 3X² + 3X + 1 can be divided by X.

Our Class 9 Maths 2.3 solutions are written by expert teachers. These teachers can explain complex things in simple language. This will help the students maintain their focus on the subjects. And the examiners too will be impressed by the student’s explanation. That is why in our Class 9 Maths Exercise 2.3 question 3 PDF, while eliminating the similar positive and negative terms, we used to red ink to prominently show the elimination.

Vedantu’s Class 9 Maths Ex 2.3 solutions are written in a clear and uncluttered way. We have not crammed up too much information in too little space. That is why, while doing the long divisions for our Ex 2.3 Class 9 Maths solutions, we used significant space so that students can understand the process without squinting their eyes or without reading multiple times.

Our NCERT Class 9 Maths Chapter 2 Exercise 2.3 solutions show step by step process of solving the problems. Not a single step has been jumped. As a result, students can understand the solutions without much head-scratching. All the steps in the long division method are there in our solutions. The examiners, too, will love the detailed answers.

Vedantu’s Class 9th Maths Chapter 2 Exercise 2.3 solutions also employ logic. In question number1 (iv), the constant is mentioned as Π (pi) instead of a or b. This should not be written as 22/7. We let Π remain as it is, just like we would have if there was a or b in its place.

Our solutions are of the highest quality. There is a very little margin of error in our solutions. The teachers who prepare these solutions are trained and experienced. You can count on our solutions. Use these answers in your exams and be sure to get good grades.

We know that this is Maths and Maths solutions should not be mugged. They should be understood. That is why we took great care while preparing these solutions - we have answered all the questions in such a way that you can understand the methods to solve the problems just by reading the PDF once. The explanatory nature of the answers will impress the examiners too.

Vedantu’s solutions are written in a simple language. You can understand the solutions without much hassle. The answers are direct and are written in a round-about manner. That is why in the solution to Question 3, after the long division method, we explained the logic behind our answer in just two short lines.

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### Learn Online With Vedantu

Vedantu’s NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Exercise 2.3 are exhaustive and easy to understand. Ye Maths solutions cannot be grasped entirely just by reading any PDF. You might want to know why the textbook only deal with terns that have one only one kind of variable in each case. What would happen if there are two variables in an equation? Can it still be called Polynomial? To banish all your confusions and to answer all your doubts, Vedantu organises online classes. These classes are conducted by teachers who are experienced, friendly and experts in their fields.

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NCERT Solutions for Class 9 Maths All Chapters

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

1. What are the topics that are covered in this Chapter?

Polynomial is an algebraic expression which includes constants, variables and exponents. It is the phrase where the variables have just positive elemental abilities. The topics that are covered in this chapter are:

• Introduction

• Polynomials in One Variable

• Zeros of Polynomials

• Remainder Theorem

• Factorisation of Polynomials

• Algebraic Identities

Polynomial expressions are algebraic equations that have two or more terms with the same variables of different exponents. This is one of the significant topics for class 9 Mathematics, that students need to learn to gain profound knowledge about complex algebraic expressions. Mathematics is one of those subjects that serve as a foundation of numbers, figures, analysis, and logic.

2. How do the solutions of Vedantu help me in securing more grades?

Our innovative learning methodology along with the smart and easy study techniques will make your learning methods more fun, interesting, interactive and in a properly planned way. Our NCERT Solutions for Class 11 Maths have been carefully designed to help you develop your knowledge base which will ultimately improve your retention rate.

All the important and necessary concepts have been covered by us in order to make your learning much easier for exam preparation. They have been crafted from the examination point of view. Our solutions are framed by our experts and they have covered each and every part of the chapter also the exercise questions which are covered at the end of the chapter.

3. Give a brief description of the topic?

Polynomial is derived from the word “poly” which means “many” and the word “nominal” which means “term”. In maths, a polynomial expression contains variables that are also known as coefficients and indeterminates. The coefficients require the operations of addition, subtraction, non-negative integer exponents of variables and multiplication.

Polynomials are utterances with one or more phrases with a non-zero coefficient. A polynomial can also have one or more than one number of terms. In the form of a polynomial, each utterance in it is known as a term.  Suppose x2 + 9x+ 4 is polynomial, then the expressions x2, 9x, and 4 are the terms of the polynomial.  Each term of the polynomial has a coefficient. For example, if 4x+1 is the polynomial, then the coefficient of x is 4.

4. What are the important terms of a polynomial?

• A term could either be a variable or a single number or it also could be a sequence of the variable with numbers.

• The degree of the polynomial is the highest power of the variable in a polynomial.

• A polynomial of degree 1 will be called as a linear polynomial.

• A polynomial of degree 2 will be called as a quadratic polynomial.

• A polynomial of degree 3 is known as a cubic polynomial.

• A polynomial of 1 term is known as a monomial.

• A polynomial of 2 terms is known as binomial.

• A polynomial of 3 terms is known as a trinomial.

• A real number ‘a’ is a zero of a polynomial p(x) if p(a) = 0, where a is also called as root of the equation p(x) = 0.

• A linear polynomial in one variable holds an individual zero, a polynomial of a non-zero constant has no zero, and every real character is a zero of the zero polynomial.

5. What concepts can I learn using the NCERT Solutions for Class 9 Maths Chapter 2?

The following concepts have been discussed in NCERT Solutions for Class 9 Maths Chapter 2:

• Introduction to polynomials

• Polynomials with One Variable

• Zeros of a Polynomial

• Remainder Theorem

• Factorisation of a Polynomial

• Identities

6. What are polynomials as discussed in Class 9 Maths Chapter 2?

The word Polynomial is derived from 2 different words. Poly means many, and Nomials meaning terms. Therefore, a Polynomial is an expression with terms that have a coefficient not equal to 0. Polynomials may have one or more terms.

7. How many questions are there in Class 9 Maths exercise 2.3?

There are three questions in Exercise 2.3 of Class 9 Maths Chapter 2. You can download NCERT Solutions for Class 9 Maths Chapter 2 Polynomials (Ex 2.3)  from the official website of Vedantu (vedantu.com). These solutions along with revision notes and study materials, are available free of cost on Vedantu (vedantu.com). You can download it using the Vedantu app as well.

8. How do I solve Class 9 Maths exercise 2.3?

Exercise 2.3 of NCERT Class 9 Maths deals with the ways in which you can use the long division method with polynomials for solving problems. To solve Class 9 Polynomials Exercise 2.3, you have to learn to divide one Polynomial with another or with a given term. This is to find the remainder of a division and also to know whether the divisor is a factor of the dividend.

You can download NCERT Solutions for Class 9 Maths Chapter 2 Polynomials (Ex 2.3)  from the official website of Vedantu (vedantu.com). These solutions along with revision notes and study materials, are available free of cost on Vedantu (vedantu.com). You can download it using the Vedantu app as well.

9. What are the most important definitions that come in Class 9 Exercise 2.3?

You have to understand a few important definitions to ace the questions from Class 9 Maths Chapter 2 Exercise 2.3. These are:

• Variables and Constants-

variable: it varies in separate cases.

Constants: it remains the same constantly in every case.

• Terms- These are the sections of any given equation.

• Polynomials- They are mathematical expressions that consist of terms.