NCERT Solutions For Class 9 Maths Chapter 2 Polynomials (Ex 2.3) Exercise 2.3

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NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.3 (Ex 2.3) (Include PDF)

The NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.3, are available at Extramarks. This chapter deals with polynomials. The NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.3, shows how to solve problems using Polynomial long division. Extramarks offers downloadable solutions for all NCERT Solutions for Class 9 Maths Chapter 2 Exercise 2.3 that have been updated in CBSE textbooks. Subjects such as Science, Mathematics, and English are easier to learn when students have access to NCERT Class 9 Solutions, NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.3, and other subject solutions available exclusively on Extramarks.

The NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.3 for Class 9 Maths Chapter 2 Exercise 2.3 covers the basics of Polynomials, including Various types of Polynomials, finding roots and solving Polynomials. A Polynomial is an algebraic expression with one or more variables. The NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.3, also cover the Remainder Theorem, Polynomial Factor Theory, Algebraic Identities, and Polynomials of Various Degrees. The NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.3, shows the difference between 1st, 2nd, and 3rd-degree Polynomials. The Remainder Theorem and the Factor Theorem are the two important sets that help identify the factors of Polynomials.

Access NCERT Solutions For Class 9 Mathematics Chapter 2 – Polynomials

Passing the CBSE exam requires good preparation and dedication. This can be achieved with the right amount of hard work and consistent practice. Working smartly and strategically  in preparation is also very important. After studying this chapter, solving Mathematics problems with NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.3 will help students understand the pattern of the exam and the weightage of each topic. Students should remember that it is important to speed up their preparation during the exam. This is the point at which many students fail to finish their assignments on time. By practising NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.3 they will acquire a tendency to think about answers to given problems at breakneck speed. There are many terms in Chapter 2 of Mathematics which are discussed in NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.3 for Class 9 Maths Ex 2.3.

Variables And Constants

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Terms

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Polynomial

The NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.3, can give the students an all-around understanding of the topics covered in this particular chapter. These solutions are easy to understand, so the students can make use of them for their own studies and make the most of them.

Examples Of Polynomials

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

In NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.3 students will learn how to divide a Polynomial by another Polynomial or any term. Class 9 Mathematics Chapter 2 Exercise 2.3 is useful for finding division remainders and also indicates whether a Polynomial is part of another Polynomial.

NCERT Solutions Class 9 Maths Chapter 2 Exercises

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Exercise 2.3 Class 9 Math act as revision notes because students have the opportunity to review all the chapters they have studied so far in a Q&A format. This is actually more beneficial than reading notes. Repeating question-and-answer patterns can help them capture or remember key points they may have missed while reading a chapter. That is, brushing and summarising the chapters for easier memorisation. So they should take a look at NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.3, and try to improve their weak areas by correcting the mistakes.

Analyze and strategize the preparation. Testing the knowledge helps them understand and reflect on how well-prepared they are. Often, students study the entire chapter well and still cannot answer the questions. This can be true for a variety of reasons, such as lack of practice, repetition, or missing important points while reading the chapter superficially. The NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.3, will help them strategize the preparation to get the maximum score from what they have already learned.

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NCERT Solutions For Class 9

Students can find the solutions to every kind of problem they face with a proper explanation.

There are six sections and five exercises in the Polynomial chapter of Class 9. However, the basic concepts in NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.3, revolve around unknown variables, calculation, and finding a solution to unknown variables. But the techniques used can be different. Practice is the key to getting the best scores on exams. However, for that, one needs proper solutions to check the steps. The NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.3, can help students get the best results and outcomes.

CBSE Study Materials For Class 9

Extramarks has done its best to provide real assistance with the NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.3 for Class 9 Polynomials. The aim was to provide enough problems and solutions for practice and to provide a strong foundation for the chapters.

There are a total of four exercises in Class 9 Polynomials, the last one being optional. For the first exercise, students need to find the roots of the polynomial p(x). For the second exercise, they have two questions. The first requires checking the relationship between the zeros and the coefficients, and the second requires finding the quadratic polynomial. The third problem has a total of 5 problems where they need to split the polynomial and find the root of the polynomial. Then there is an optional 5-question exercise where they need to find the roots of a polynomial.

A compact exercise follows each topic. This exercise aims to test your knowledge and deep understanding of the various theorems and concepts presented in this chapter. Nevertheless, it should be noted that the numerical problems in this chapter are mainly based on certain theorems and other related concepts. Numerous examples of solving numerical problems are also provided to deepen the understanding of these topics and related concepts. Additionally, detailed step-by-step instructions are provided for each solved example. It helps to understand the methods used to tackle different types of questions in order to solve them accurately.

CBSE Study Materials

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Q.1  Find the remainder when x3 – ax2 + 6x – a is divided by x – a.

Ans

${\text{x}}^{\text{3}}–{\text{ax}}^{\text{2}}+\text{6x}–\text{a is divided by x}–\text{a}$ 

$\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{x}^{2}+6\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{x}–\text{a}\overline{){\text{x}}^{\text{3}}–{\text{ax}}^{\text{2}}+\text{6x}–\text{a}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\underset{¯}{±{x}^{3}\mp a{x}^{2}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}6x-a\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\underset{¯}{\text{\hspace{0.17em}}±6x\mp 6a}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\underset{¯}{\text{\hspace{0.17em}}\text{\hspace{0.17em}}5a\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}}\\ \end{array}$

T
hus, the denominator is 5a.

$\begin{array}{l}Another\text{method:}\\ \text{Let}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{p}\left(x\right)={\text{x}}^{\text{3}}–{\text{ax}}^{\text{2}}+\text{6x}–\text{a and the zero of x}–\text{a is a}\text{.}\\ \text{So, p}\left(a\right)={\left(a\right)}^{\text{3}}–{\left(a\right)}^{\text{3}}+\text{6}\left(a\right)–\text{a}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=5a\\ \text{Thus, the remainder is}5a\text{.}\end{array}$

Q.2 Check whether 7 + 3x is a factor of 3x3 + 7x.

Ans

$\begin{array}{l}\text{Let}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{p}\left(x\right)={\text{3x}}^{\text{3}}+\text{7x is divided by 7}+\text{3x i}\text{.e}\text{., x=}-\frac{7}{3}\mathrm{if}7+3\mathrm{x}=0\text{.}\\ \text{So, p}\left(-\frac{7}{3}\right)=\text{3}{\left(-\frac{7}{3}\right)}^{\text{3}}+\text{7}\left(-\frac{7}{3}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=-\frac{343}{9}-\frac{49}{3}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{-343-147}{9}=-\frac{490}{9}\\ \text{Thus, the remainder is}-\frac{490}{9}\text{.}\\ \mathrm{Therefore},\text{7}+{\text{3x is not a factor of 3x}}^{\text{3}}+\text{7x}.\end{array}$

Q.3

$\begin{array}{l}\text{Determine which of the following polynomials has}\left(\text{x+1}\right)\text{a}\\ \text{factor:}\\ \left(\text{i}\right){\text{ x}}^{\text{3}}{\text{+x}}^{\text{2}}\text{+x+1}\\ \left(\text{ii}\right){\text{ x}}^{\text{4}}{\text{+x}}^{\text{3}}{\text{+x}}^{\text{2}}\text{+x+1}\\ \left(\text{iii}\right){\text{ x}}^{\text{4}}{\text{+3x}}^{\text{3}}{\text{+3x}}^{\text{2}}\text{+x+1}\\ \left(\text{iv}\right){\text{ x}}^{\text{3}}-{\text{x}}^{\text{2}}-\left(\text{2+}\sqrt{\text{2}}\right)\text{x+}\sqrt{\text{2}}\end{array}$

Ans

(i) Let P(x) = x3 + x2 + x + 1 is divided by x+1 i.e., x = –1 if x + 1 = 0.

So, P(–1) = (–1)3 + (–1)2 + (–1) + 1

= –1 + 1 –1 + 1 = 0

Since, remainder is 0. So, (x + 1) is a factor of the given polynomial.

(ii) Let P(x) = x4 + x3 + x2 + x + 1 is divided by x+1
i.e., x = –1 if x + 1 = 0.

So, P(–1) = (–1)4 + (–1)3 + (–1)2 + (–1) + 1

= 1 – 1 + 1 – 1 + 1 = 1

Since, remainder is 1. So, (x + 1) is not a factor of the given polynomial.
(iii) Let P(x) = x4 + 3x3 + 3x2 + x + 1 is divided by x+1
i.e., x = –1 if x + 1 = 0.

So, P(–1) = (–1)4 + 3(–1)3 + 3(–1)2 + (–1) + 1

= 1 – 3 + 3 – 1 + 1 = 1

Since, remainder is 1. So, (x + 1) is not a factor of the given polynomial.

$\begin{array}{l}\left(\mathrm{iv}\right)\mathrm{}\text{Let P}\left(\text{x}\right)=\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{x}}^{3}-{\mathrm{x}}^{2}-\left(2+\sqrt{2}\right)\mathrm{x}+\sqrt{2}\text{is divided by x}+\text{1 i}.\text{e}.,\text{x =}–\text{1 if x + 1 = 0}.\\ \text{So},\text{P}\left(–\text{1}\right)\text{}={\left(–\text{1}\right)}^{3}-{\left(–\text{1}\right)}^{2}-\left(2+\sqrt{2}\right)\left(–\text{1}\right)+\sqrt{2}\\ \text{\hspace{0.17em}\hspace{0.17em}}=-\text{1}-\text{1}+\text{2}+\sqrt{2}\text{}+\text{}\sqrt{2}\text{}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\text{2}\sqrt{2}\ne 0\\ \text{Since},\text{remainder is 2}\sqrt{2}.\text{So},\text{}\left(\text{x}+\text{1}\right)\text{is not a factor of}\mathrm{}\mathrm{the}\text{given polynomial}.\end{array}$