NCERT Solutions Class 8 Maths Chapter 14 Exercise 14.3

NCERT Solutions for Class 8 Mathematics Chapter 14 Exercise 14.3 are designed to help students master the concepts and gain a thorough knowledge of the different types of NCERT Solutions for Class 8 Mathematics Chapter 14 Exercise 14.3 questions that appear in the exam. All questions have corresponding solutions in NCERT Solutions for Class 8 Mathematics Chapter 14 Exercise 14.3 to help students revise CBSE Class 8 Mathematics syllabus. NCERT Solutions for Class 8 Mathematics Chapter 14 Exercise 14.3 are developed by highly qualified subject matter experts according to the CBSE syllabus. NCERT Solutions for Class 8 Mathematics Chapter 14 Exercise 14.3 Mathematics helps students streamline their last-minute corrections. Students can download the NCERT Solutions for Class 8 Mathematics Chapter 14 Exercise 14.3 PDF to study offline.

NCERT Solutions for Class 8 Mathematics Chapter 14 Exercise 14.3 is an important topic covered in the syllabus. From an exam perspective, this is an assessment chapter, so students should focus on NCERT Solutions for Class 8 Mathematics Chapter 14 Exercise 14.3 to do well on the exam.

Factorisation formulas may seem complicated at first, but they become easier with practice. NCERT Solutions for Class 8 Mathematics Chapter 14 Exercise 14.3 will help students cover this chapter easily.

NCERT Solutions for Class 8 Mathematics Chapter 14 Factorisation (EX 14.3) Exercise 14.3

NCERT Solutions for Class 8 Mathematics, Chapter 14, Exercise 14.3, covers all topics and subtopics related to factoring and its applications. NCERT Solutions for Class 8 Mathematics, Chapter 14, Exercise 14.3, is a critical skill required for advanced mathematics solutions. Learning to factor expressions is a complex process that requires a deep, step-by-step understanding of factors and how to list them. By thoroughly practising NCERT Solutions for Class 8 Mathematics Chapter 14 Exercise 14.3, students can acquire this knowledge quickly and prepare for the exam. These NCERT Solutions for Class 8 Mathematics Chapter 14 Exercise 14.3 solutions  instill confidence in students as they prepare them for a variety of competitive studies. These NCERT Solutions for Class 8 Mathematics Chapter 14 Exercise 14.3 are well-structured resources that are highly skilled at building a strong mathematical foundation. By thoroughly practising all the questions available in these NCERT Solutions for Class 8 Mathematics Chapter 14 Exercise 14.3 solutions, students can easily explore interesting ways to learn this topic. The series of questions provided in NCERT Solutions for Class 8 Mathematics Chapter 14, Exercise 14.3, promotes students’ problem-solving skills.

NCERT Solutions for Class 8 Mathematics Chapter 14 Exercise 14.3 and answers help students find the perfect answer for math exams. These NCERT Solutions for Class 8 Mathematics Chapter 14 Exercise 14.3 exercises can provide students with topic-related preparation strategies. Important topics introduced in NCERT Solutions for Class 8 Mathematics Chapter 14 Exercise 14.3 include factorization, factors of natural numbers, factors of algebraic expressions, and division of algebraic expressions.

NCERT response to NCERT Solutions for Class 8 Mathematics Chapter 14 Exercise 14.3:

Students can get detailed solutions on NCERT Solutions for Class 8 Mathematics Chapter 14 Exercise 14.3 to all questions in the following exercises.

Exercise 14.1 Solution: 3 short-answer questions

It is primarily concerned with algebraic expressions using factorization and algebraic identities. Students will also learn how to divide algebraic expressions, how to divide a monomial by another monomial, how to divide a polynomial by a monomial, and how to divide a polynomial by a polynomial.

Students can download the solutions in PDF format. These printable files can be taken anywhere, anytime, and practised with friends, so students have physical copies for uninterrupted learning time. They may also be stored and used during exams and revised during participation in the Olympiad and competitive exams.

Learning the key methods and algebraic formulas of prime factorization is an important skill. Mastering this subject will enable students to perform well in exams. It includes good examples of all necessary factorization methods. Here are some ways to factorise the algebraic expressions described in these NCERT Solutions for Class 8 Mathematics Chapter 14 Exercise 14.3 solutions.

Common Factor Method: This method allows the expression to be expressed as a product of terms. For example: ax^2 +axy = ax(x+y).

Regrouping Factors: Regrouping the terms of an expression allows the expression to be expressed as the product of its factors. For example: a2 + bc + ab + ac = (a + b) (a + c). Factoring by Identities: Factoring by identities is a simple way to factor algebraic expressions. For example, a2 + 2ab + b2 = (a + b)2

Factors of the form (a+x)(b+x): These expressions are not exact squares and can be factored using this form. x2+ (a + b) x + ab = (x + a) (x + b)

Access NCERT Solutions for Class 8 Mathematics Chapter 14 : Factorization

Solutions are written according to the NCERT textbooks to provide a comprehensive study of the entire syllabus. The simplified format of these NCERT Solutions for Class 8 Mathematics Chapter 14 Exercise 14.3 solutions makes it easier for students to learn and explore this topic. Cleverly placed exercises and sample tasks provide strategic learning of individual concepts. Students can go to the link on  NCERT Solutions for Class 8 Mathematics Chapter 14 Exercise 14.3 to practise NCERT Solutions for Class 8 Mathematics Factorization.

Class 8 Mathematics Chapter 14 Ex 14.1 – 3 Questions

Class 8 Mathematics Chapter 14 Ex 14.2 – 5 Questions

Class 8 Mathematics Chapter 14 Ex 14.3 – 5 Questions

Class 8 Mathematics Chapter 14 Ex 14.4 – 20 Questions

Topics covered: NCERT Solutions for Class 8 Mathematics Chapter 14 Exercise 14.3 covers all important topics related to factorization, including factors of natural numbers, factors of algebraic expressions, methods of factoring, and division of algebraic expressions.

Total Questions: Class 8 Mathematics, Chapter 14, Factoring has 33 questions, mostly short-answer questions with subparts. These problems are mainly based on factoring algebraic expressions and dividing polynomials by polynomials.

On this page in PDF format, choosing these solutions is considered the best option for CBSE students when it comes to exam preparation. Students can download this solution at their convenience or study directly online from Extramarks’ website/app.

Extramarks’ subject matter experts solved the exercise problems and questions meticulously and in compliance with all the CBSE guidelines. A Class 8 student who is familiar with every concept in a math textbook and every problem from the exercises in it will easily get the best possible score in the final exam. Using these solutions for Class 8 Mathematics Chapter 14 Exercise 14.3, students can easily understand the patterns of questions that may appear in exams andcan prepare according to the final exam.

14.1 Introduction

The first section of each chapter is almost always the same. This is the introductory part. Here, the introduction can remind students of the elements they have learned in previous grades. Students have already learned how to find the meaning of divisors of natural numbers and divisors of algebraic expressions. Next, in this chapter, students will learn more about factoring large expressions using different methods.

14.2 What is factorization?

Chapter 12 Factorization PDF in Class 8 mathematics explains the importance of factorization. Factorization is the process by which a given expression is expressed as the product of two factors. These factors can be either numbers, variables, or algebraic expressions themselves. The PDF also explains that there are several systematic approaches to finding the factors of a given expression. Let’s take a look at the methods for determining the factors.

14.2.1 Common factor method

This is the first way to find the factors of a given expression. According to NCERT Solutions Class 8 Factorization, a given formula can be expressed as a product of terms. This becomes clear when looking at two or more examples in the PDF.

14.2.2 Factorization by Reordering Terms

The next section of this chapter deals with alternative methods of factorization. Chapter 14, Class 8, explains that in finding the factors of a given expression, students assume that a single term is common and consider it the common factor. It can be done by forming a group of two. for example,

6 𝑝 − 12 𝑞

= 6𝑝 − (6 × 2) 𝑞

= 6(𝑝 − 2𝑞) (usually taken as 6)

Students will use three short-answer questions for Exercise 14.1.

14.2.3 Factorization

Class 8 Mathematics Chapter 14 Exercise 14.3 arrived at the third factorization method in this section. Here, students must find identical terms and then isolate the terms that directly represent the factors.

14.2.4 Factors of the form (x+a)(x+b)

This is the fourth way to solve factorizations. NCERT Solutions explains that this method is usually used when there is no perfect square in the expression. Although suitable for perfect squares, this method can also be used to get the factors of a given expression without using perfect squares. There are different ways to find the factor, but the solution is the same; the steps may differ.

Exercise 14.2 has five short-answer problems, each with a set of equations to solve.

14.3 Division of algebraic expressions

NCERT Solutions for Class 8 covers advanced factorisation concepts. After students learn to add, subtract, and multiply algebraic expressions, NCERT Solutions for Class 8 Mathematics Chapter 14 Exercise 14.3  will help them understand the division of two or more expressions. But it has two sub-concepts. This is because it can be classified according to the degree of expression. From the notes in Class 8 Mathematics Exercise 14.3, students will learn:

Dividing a monomial by another monomial.

Division of a polynomial by a monomial.

NCERT Solutions for Class 8 Mathematics Chapter 14 Factorisation Exercise 14.3

In addition to these NCERT Solutions for Class 8 Mathematics Chapter 14 Exercise 14.3, there are many exercises in this chapter with countless questions. As already mentioned, all these questions are resolved or answered by technical experts. Therefore, they should all be of the highest quality and can be referenced by anyone while preparing for the exam. Understanding all the concepts in the textbook and solving the adjacent exercise problems is very important for getting the best grades in the class.

Solutions are written to be comprehensive and easy to understand.

Below are detailed answers to each problem covered in the NCERT Solutions for Class 8 Mathematics Chapter 14 Exercise 14.3.

Subject matter experts create NCERT Solutions for Class 8 Mathematics Chapter 14 Exercise 14.3 so that people get accurate, step-by-step solutions.

NCERT Solutions for Class 8 Mathematics Chapter 14 Exercise 14.3 is available for download in PDF format. Class 8 Mathematics Exercise 14.3 Solution focuses on clarifying concepts by providing insight into NCERT Solutions for Class 8 Mathematics Chapter 14 Exercise 14.3 to help prepare for the exam.

Students can download the NCERT Solutions for Class 8 Mathematics Chapter 14 Exercise 14.3 from the Extramarks website and get ready for the exam. If students have the Extramarks app installed on their smartphone, they can also download it from the Extramarks app.

NCERT Solutions for Class 8

Key Features of the NCERT Solutions for Class 8 Mathematics Chapter 14 Exercise 14.3

These NCERT Solutions for Class 8 Mathematics Chapter 14 Exercise 14.3 help students understand concepts clearly.

NCERT Class 8 Mathematics Chapter 14 Exercise 14.3 uses simple and concise language to describe topics.

All concepts in the solutions are explained in detail.

A subject matter expert has put together all her exercises in one place for students to practice.

It helps to prepare for competitive exams.

Q.1

$\begin{array}{l}\text{Carry out the following divisions.}\\ {\text{(i) 28x}}^{\text{4}}\text{÷ 56x}\\ \text{(ii) }-{\text{36y}}^{\text{3}}{\text{÷ 9y}}^{\text{2}}\text{}\\ {\text{(iii) 66pq}}^{\text{2}}{\text{r}}^{\text{3}}{\text{÷ 11qr}}^{\text{2}}\\ {\text{(iv) 34x}}^{\text{3}}{\text{y}}^{\text{3}}{\text{z}}^{\text{3}}{\text{÷ 51xy}}^{\text{2}}{\text{z}}^{\text{3}}\text{}\\ {\text{(v) 12a}}^{\text{8}}{\text{b}}^{\text{8}}\text{÷ (}-{\text{6a}}^{\text{6}}{\text{b}}^{\text{4}}\text{)}\end{array}$

Ans

$\begin{array}{l}\left(\mathrm{i}\right)\text{ }28{\mathrm{x}}^{4}÷56\mathrm{x}\\ 28{\mathrm{x}}^{4}=2×2×7×\mathrm{x}×\mathrm{x}×\mathrm{x}×\mathrm{x}\\ 56\mathrm{x}\text{\hspace{0.17em}}=2×2×2×7×\mathrm{x}\\ \\ \therefore \frac{28{\mathrm{x}}^{4}}{56\mathrm{x}}\\ =\frac{\overline{)2}×\overline{)2}×\overline{)7}×\mathrm{x}×\mathrm{x}×\mathrm{x}×\overline{)\mathrm{x}}}{\overline{)2}×\overline{)2}×2×\overline{)7}×\overline{)\mathrm{x}}}\\ =\frac{{\mathrm{x}}^{3}}{2}\\ \\ \left(\mathrm{ii}\right)-36{\mathrm{y}}^{3}÷9{\mathrm{y}}^{2}\\ 36{\mathrm{y}}^{3}=2×2×3×3×\mathrm{y}×\mathrm{y}×\mathrm{y}\\ 9{\mathrm{y}}^{2}\text{}=3×3×\mathrm{y}×\mathrm{y}\\ \\ \therefore \frac{-36{\mathrm{y}}^{3}}{9{\mathrm{y}}^{2}}\\ =\frac{-\overline{)3}×\overline{)3}×2×2×\mathrm{y}×\overline{)\mathrm{y}}×\overline{)\mathrm{y}}}{\overline{)3}×\overline{)3}×\overline{)\mathrm{y}}×\overline{)\mathrm{y}}}\\ =-4\mathrm{y}\\ \\ \left(\mathrm{iii}\right)66{\mathrm{pq}}^{2}{\mathrm{r}}^{3}÷11{\mathrm{qr}}^{2}\\ 66{\mathrm{pq}}^{2}{\mathrm{r}}^{3}=2×3×11×\mathrm{p}×\mathrm{q}×\mathrm{q}×\mathrm{r}×\mathrm{r}×\mathrm{r}\\ 11{\mathrm{qr}}^{2}\text{}=11×\mathrm{q}×\mathrm{r}×\mathrm{r}\\ \\ \therefore \frac{66{\mathrm{pq}}^{2}{\mathrm{r}}^{3}}{11{\mathrm{qr}}^{2}}\\ =\frac{\overline{)11}×3×2×\mathrm{p}×\overline{)\mathrm{q}}×\mathrm{q}×\overline{)\mathrm{r}×\mathrm{r}}×\mathrm{r}}{\overline{)11}×\overline{)\mathrm{q}}×\overline{)\mathrm{r}×\mathrm{r}}}\\ =6\mathrm{pqr}\end{array}$

$\begin{array}{l}\left(\mathrm{iv}\right)\text{ }34{\mathrm{x}}^{3}{\mathrm{y}}^{3}{\mathrm{z}}^{3}÷51{\mathrm{xy}}^{2}{\mathrm{z}}^{3}\\ 34{\mathrm{x}}^{3}{\mathrm{y}}^{3}{\mathrm{z}}^{3}=2×17×\mathrm{x}×\mathrm{x}×\mathrm{x}×\mathrm{y}×\mathrm{y}×\mathrm{y}×\mathrm{z}×\mathrm{z}×\mathrm{z}\\ 51{\mathrm{xy}}^{2}{\mathrm{z}}^{3}\text{}=3×17×\mathrm{x}×\mathrm{y}×\mathrm{y}×\mathrm{z}×\mathrm{z}×\mathrm{z}\\ \therefore \frac{34{\mathrm{x}}^{3}{\mathrm{y}}^{3}{\mathrm{z}}^{3}}{51{\mathrm{xy}}^{2}{\mathrm{z}}^{3}}\\ =\frac{2×\overline{)17}×\mathrm{x}×\overline{)\mathrm{x}}×\mathrm{x}×\overline{)\mathrm{y}}×\overline{)\mathrm{y}}×\mathrm{y}×\overline{)\mathrm{z}}×\overline{)\mathrm{z}}×\overline{)\mathrm{z}}}{3×\overline{)17}×\overline{)\mathrm{x}}×\overline{)\mathrm{y}}×\overline{)\mathrm{y}}×\overline{)\mathrm{z}}×\overline{)\mathrm{z}}×\overline{)\mathrm{z}}}\\ =\frac{2{\mathrm{x}}^{2}\mathrm{y}}{3}\\ \\ \left(\mathrm{v}\right)\text{ }12{\mathrm{a}}^{8}{\mathrm{b}}^{8}÷\left(-6{\mathrm{a}}^{6}{\mathrm{b}}^{4}\right)\\ 12{\mathrm{a}}^{8}{\mathrm{b}}^{8}=2×2×3×\mathrm{a}×\mathrm{a}×\mathrm{a}×\mathrm{a}×\mathrm{a}×\mathrm{a}×\mathrm{a}×\mathrm{a}×\mathrm{b}×\mathrm{b}×\mathrm{b}×\mathrm{b}×\mathrm{b}×\mathrm{b}×\mathrm{b}×\mathrm{b}\\ 6{\mathrm{a}}^{6}{\mathrm{b}}^{4}\text{}=3×2×\mathrm{a}×\mathrm{a}×\mathrm{a}×\mathrm{a}×\mathrm{a}×\mathrm{a}×\mathrm{b}×\mathrm{b}×\mathrm{b}×\mathrm{b}\\ \therefore \frac{12{\mathrm{a}}^{8}{\mathrm{b}}^{8}}{\left(-6{\mathrm{a}}^{6}{\mathrm{b}}^{4}\right)}\\ =\frac{2×\overline{)2×3}×\overline{)\mathrm{a}}×\overline{)\mathrm{a}}×\overline{)\mathrm{a}}×\overline{)\mathrm{a}}×\overline{)\mathrm{a}}×\overline{)\mathrm{a}}×\mathrm{a}×\mathrm{a}×\overline{)\mathrm{b}}×\overline{)\mathrm{b}}×\overline{)\mathrm{b}}×\overline{)\mathrm{b}}×\mathrm{b}×\mathrm{b}×\mathrm{b}×\mathrm{b}}{-\overline{)3×2}×\overline{)\mathrm{a}}×\overline{)\mathrm{a}}×\overline{)\mathrm{a}}×\overline{)\mathrm{a}}×\overline{)\mathrm{a}}×\overline{)\mathrm{a}}×\overline{)\mathrm{b}}×\overline{)\mathrm{b}}×\overline{)\mathrm{b}}×\overline{)\mathrm{b}}}\\ =-2{\mathrm{a}}^{2}{\mathrm{b}}^{4}\end{array}$

Q.2

$\begin{array}{l}\text{Divide the given polynomial by the given monomial.}\\ {\text{(i) (5x}}^{\text{2}}\text{ }-\text{ 6x) ÷ 3x}\\ \left(\mathrm{ii}\right)\left(3{\mathrm{y}}^{\text{8}}\text{ }-{\text{ 4y}}^{\text{6}}{\text{+ 5y}}^{\text{4}}{\text{) ÷ y}}^{\text{4}}\\ {\text{(iii) 8(x}}^{\text{3}}{\text{y}}^{\text{2}}{\text{z}}^{\text{2}}{\text{+ x}}^{\text{2}}{\text{y}}^{\text{3}}{\text{z}}^{\text{2}}{\text{+ x}}^{\text{2}}{\text{y}}^{\text{2}}{\text{z}}^{\text{3}}{\text{) ÷ 4x}}^{\text{2}}{\text{y}}^{\text{2}}{\text{z}}^{\text{2}}\\ {\text{(iv) (x}}^{\text{3}}{\text{+ 2x}}^{\text{2}}\text{+ 3x) ÷ 2x}\\ {\text{(v) (p}}^{\text{3}}{\text{q}}^{\text{6}}\text{ }-{\text{ p}}^{\text{6}}{\text{q}}^{\text{3}}{\text{) ÷ p}}^{\text{3}}{\text{q}}^{\text{3}}\end{array}$

Ans

$\begin{array}{l}\\ \left(\mathrm{i}\right)\text{ }\left(5{\mathrm{x}}^{2}-6\mathrm{x}\right)÷3\mathrm{x}\\ \frac{\left(5{\mathrm{x}}^{2}-6\mathrm{x}\right)}{3\mathrm{x}}\\ =\frac{\overline{)\mathrm{x}}\left(5\mathrm{x}-6\right)}{3\overline{)\mathrm{x}}}\\ =\frac{\left(5\mathrm{x}-6\right)}{3}\\ \\ \left(\mathrm{ii}\right)\text{ }\left(3{\mathrm{y}}^{8}-4{\mathrm{y}}^{6}+5{\mathrm{y}}^{4}\right)÷{\mathrm{y}}^{4}\\ \frac{\left(3{\mathrm{y}}^{8}-4{\mathrm{y}}^{6}+5{\mathrm{y}}^{4}\right)}{{\mathrm{y}}^{4}}\\ =\frac{\overline{){\mathrm{y}}^{4}}\left(3{\mathrm{y}}^{4}-4{\mathrm{y}}^{2}+5\right)}{\overline{){\mathrm{y}}^{4}}}\\ =3{\mathrm{y}}^{4}-4{\mathrm{y}}^{2}+5\end{array}$

$\begin{array}{l}\left(\mathrm{iii}\right)\text{ }8\left({\mathrm{x}}^{3}{\mathrm{y}}^{2}{\mathrm{z}}^{2}+{\mathrm{x}}^{2}{\mathrm{y}}^{3}{\mathrm{z}}^{2}+{\mathrm{x}}^{2}{\mathrm{y}}^{2}{\mathrm{z}}^{3}\right)÷4{\mathrm{x}}^{2}{\mathrm{y}}^{2}{\mathrm{z}}^{2}\\ \\ \frac{8\left({\mathrm{x}}^{3}{\mathrm{y}}^{2}{\mathrm{z}}^{2}+{\mathrm{x}}^{2}{\mathrm{y}}^{3}{\mathrm{z}}^{2}+{\mathrm{x}}^{2}{\mathrm{y}}^{2}{\mathrm{z}}^{3}\right)}{4{\mathrm{x}}^{2}{\mathrm{y}}^{2}{\mathrm{z}}^{2}}\\ =\frac{8\overline{){\mathrm{x}}^{2}{\mathrm{y}}^{2}{\mathrm{z}}^{2}}\left(\mathrm{x}+\mathrm{y}+\mathrm{z}\right)}{4\overline{){\mathrm{x}}^{2}{\mathrm{y}}^{2}{\mathrm{z}}^{2}}}\\ =2\left(\mathrm{x}+\mathrm{y}+\mathrm{z}\right)\end{array}$ $\begin{array}{l}\end{array}$

$\begin{array}{l}\left(\mathrm{iv}\right)\text{ }\left({\mathrm{x}}^{3}+2{\mathrm{x}}^{2}+3\mathrm{x}\right)÷2\mathrm{x}\\ \\ \frac{\left({\mathrm{x}}^{3}+2{\mathrm{x}}^{2}+3\mathrm{x}\right)}{2\mathrm{x}}\\ =\frac{\overline{)\mathrm{x}}\left({\mathrm{x}}^{2}+2\mathrm{x}+3\right)}{2\overline{)\mathrm{x}}}\\ =\frac{\left({\mathrm{x}}^{2}+2\mathrm{x}+3\right)}{2}\\ \\ \left(\mathrm{v}\right)\text{ }\left({\mathrm{p}}^{3}{\mathrm{q}}^{6}-{\mathrm{p}}^{6}{\mathrm{q}}^{3}\right)÷{\mathrm{p}}^{3}{\mathrm{q}}^{3}\\ \\ \frac{{\mathrm{p}}^{3}{\mathrm{q}}^{6}-{\mathrm{p}}^{6}{\mathrm{q}}^{3}}{{\mathrm{p}}^{3}{\mathrm{q}}^{3}}\\ =\frac{\overline{){\mathrm{p}}^{3}{\mathrm{q}}^{3}}\left({\mathrm{q}}^{3}-{\mathrm{p}}^{3}\right)}{\overline{){\mathrm{p}}^{3}{\mathrm{q}}^{3}}}\\ ={\mathrm{q}}^{3}-{\mathrm{p}}^{3}\end{array}$

Q.3 Work out the following divisions.

(i) (10x – 25) ÷ 5

(ii) (10x – 25) ÷ (2x – 5)

(iii) 10y(6y + 21) ÷ 5(2y + 7)

(iv) 9x2y2(3z – 24) ÷ 27xy(z – 8)

(v) 96abc(3a – 12) (5b – 30) ÷ 144(a – 4) (b – 6)

Ans

$\begin{array}{l}\left(\mathrm{i}\right)\mathrm{}\left(10\mathrm{x}–25\right)÷5\\ \frac{\mathrm{}\left(10\mathrm{x}–25\right)}{5}\\ =\frac{\mathrm{}5\left(2\mathrm{x}–5\right)}{5}\\ =2\mathrm{x}–5\\ \\ \left(\mathrm{ii}\right)\left(10\mathrm{x}–25\right)÷\left(2\mathrm{x}–5\right)\\ \frac{\left(10\mathrm{x}–25\right)}{\left(2\mathrm{x}–5\right)}\\ =\frac{5\left(2\mathrm{x}–5\right)}{\left(2\mathrm{x}–5\right)}\\ =5\end{array}$

$\begin{array}{l}\left(\mathrm{iii}\right)\mathrm{}10\mathrm{y}\left(6\mathrm{y}+21\right)÷5\left(2\mathrm{y}+7\right)\\ \frac{10\mathrm{y}\left(6\mathrm{y}+21\right)}{5\left(2\mathrm{y}+7\right)}\\ =\frac{10\mathrm{y}×3\overline{)\left(2\mathrm{y}+7\right)}}{5\overline{)\left(2\mathrm{y}+7\right)}}\\ =6\mathrm{y}\\ \\ \left(\mathrm{iv}\right)\mathrm{}9{\mathrm{x}}^{2}{\mathrm{y}}^{2}\left(3\mathrm{z}–24\right)÷27\mathrm{xy}\left(\mathrm{z}–8\right)\\ \frac{9{\mathrm{x}}^{2}{\mathrm{y}}^{2}\left(3\mathrm{z}–24\right)}{27\mathrm{xy}\left(\mathrm{z}–8\right)}\\ =\frac{9{\mathrm{x}}^{2}{\mathrm{y}}^{2}×3\overline{)\left(\mathrm{z}–8\right)}}{27\mathrm{xy}\overline{)\left(\mathrm{z}–8\right)}}\\ =\mathrm{xy}\end{array}$ $\begin{array}{l}\end{array}$

$\begin{array}{l}\left(\mathrm{v}\right)96\mathrm{abc}\left(3\mathrm{a}–12\right)\left(5\mathrm{b}–\text{}30\right)÷144\left(\mathrm{a}–4\right)\left(\mathrm{b}–6\right)\\ \frac{96\mathrm{abc}\left(3\mathrm{a}–12\right)\left(5\mathrm{b}–\text{}30\right)}{144\left(\mathrm{a}–4\right)\left(\mathrm{b}–6\right)}\\ =\frac{96\mathrm{abc}×3\overline{)\left(\mathrm{a}–4\right)}×5\overline{)\left(\mathrm{b}–\text{6}\right)}}{144\overline{)\left(\mathrm{a}–4\right)}\overline{)\left(\mathrm{b}–6\right)}}\\ =10\mathrm{abc}\end{array}$

Q.4

$\begin{array}{l}\text{Divide as directed.}\\ \text{(i) 5(2x + 1) (3x + 5) ÷ (2x + 1)}\\ \text{(ii) 26xy(x + 5) (y}-\text{ 4) ÷ 13x(y – 4)}\\ \text{(iii) 52pqr (p + q) (q + r) (r + p) ÷ 104pq(q + r) (r + p)}\\ {\text{(iv) 20(y + 4) (y}}^{\text{2}}\text{+ 5y + 3) ÷ 5(y + 4)}\\ \text{(v) x(x + 1) (x + 2) (x + 3) ÷ x(x + 1)}\end{array}$

Ans

$\begin{array}{l}\left(\mathrm{i}\right)\text{ }5\left(2\mathrm{x}+1\right)\left(3\mathrm{x}+5\right)÷\left(2\mathrm{x}+1\right)\\ \frac{5\left(2\mathrm{x}+1\right)\left(3\mathrm{x}+5\right)}{\left(2\mathrm{x}+1\right)}\\ =\frac{5\overline{)\left(2\mathrm{x}+1\right)}\left(3\mathrm{x}+5\right)}{\overline{)\left(2\mathrm{x}+1\right)}}\\ =5\left(3\mathrm{x}+5\right)\\ \\ \left(\mathrm{ii}\right)\text{ }26\mathrm{xy}\left(\mathrm{x}+5\right)\left(\mathrm{y}-4\right)÷13\mathrm{x}\left(\mathrm{y}-4\right)\\ \frac{26\mathrm{xy}\left(\mathrm{x}+5\right)\left(\mathrm{y}-4\right)}{13\mathrm{x}\left(\mathrm{y}-4\right)}\\ =\frac{26\overline{)\mathrm{x}}\mathrm{y}\left(\mathrm{x}+5\right)\overline{)\left(\mathrm{y}-4\right)}}{13\overline{)\mathrm{x}}\overline{)\left(\mathrm{y}-4\right)}}\\ =2\mathrm{y}\left(\mathrm{x}+5\right)\end{array}$

$\begin{array}{l}\left(\mathrm{iii}\right)\text{ }52\mathrm{pqr}\left(\mathrm{p}+\mathrm{q}\right)\left(\mathrm{q}+\mathrm{r}\right)\left(\mathrm{r}+\mathrm{p}\right)÷104\mathrm{pq}\left(\mathrm{q}+\mathrm{r}\right)\left(\mathrm{r}+\mathrm{p}\right)\\ \\ \frac{52\mathrm{pqr}\left(\mathrm{p}+\mathrm{q}\right)\left(\mathrm{q}+\mathrm{r}\right)\left(\mathrm{r}+\mathrm{p}\right)}{104\mathrm{pq}\left(\mathrm{q}+\mathrm{r}\right)\left(\mathrm{r}+\mathrm{p}\right)}\end{array}$ $\begin{array}{l}\end{array}$

$\begin{array}{l}=\frac{\overline{)2×2}×\overline{)13}×\overline{)\mathrm{pq}}\mathrm{r}\left(\mathrm{p}+\mathrm{q}\right)\overline{)\left(\mathrm{q}+\mathrm{r}\right)}\overline{)\left(\mathrm{r}+\mathrm{p}\right)}}{\overline{)2×2}×2×\overline{)13}×\overline{)\mathrm{pq}}\overline{)\left(\mathrm{q}+\mathrm{r}\right)}\overline{)\left(\mathrm{r}+\mathrm{p}\right)}}\\ =\frac{\mathrm{r}\left(\mathrm{p}+\mathrm{q}\right)}{2}\end{array}$ $\begin{array}{l}\end{array}$

$\begin{array}{l}\left(\mathrm{iv}\right)\text{ }20\left(\mathrm{y}+4\right)\left({\mathrm{y}}^{2}+5\mathrm{y}+3\right)÷5\left(\mathrm{y}+4\right)\\ \\ \frac{20\left(\mathrm{y}+4\right)\left({\mathrm{y}}^{2}+5\mathrm{y}+3\right)}{5\left(\mathrm{y}+4\right)}\\ =\frac{\overline{)5}×4\overline{)\left(\mathrm{y}+4\right)}\left({\mathrm{y}}^{2}+5\mathrm{y}+3\right)}{\overline{)5}\overline{)\left(\mathrm{y}+4\right)}}\\ =4\left({\mathrm{y}}^{2}+5\mathrm{y}+3\right)\end{array}$ $\begin{array}{l}\end{array}$

$\begin{array}{l}\left(\mathrm{v}\right)\text{ }\mathrm{x}\left(\mathrm{x}+1\right)\left(\mathrm{x}+2\right)\left(\mathrm{x}+3\right)÷\mathrm{x}\left(\mathrm{x}+1\right)\\ \\ \frac{\mathrm{x}\left(\mathrm{x}+1\right)\left(\mathrm{x}+2\right)\left(\mathrm{x}+3\right)}{\mathrm{x}\left(\mathrm{x}+1\right)}\\ =\frac{\overline{)\mathrm{x}\left(\mathrm{x}+1\right)}\left(\mathrm{x}+2\right)\left(\mathrm{x}+3\right)}{\overline{)\mathrm{x}\left(\mathrm{x}+1\right)}}\\ =\left(\mathrm{x}+2\right)\left(\mathrm{x}+3\right)\end{array}$

Q.5

$\begin{array}{l}\text{Factorise the expressions and divide them as directed.}\\ {\text{(i) (y}}^{\text{2}}\text{+ 7y + 10) ÷ (y + 5)}\\ {\text{(ii) (m}}^{\text{2}}\text{ }-\text{ 14m }-\text{ 32) ÷ (m + 2)}\\ {\text{(iii) (5p}}^{\text{2}}\text{ }-\text{ 25p + 20) ÷ (p }-\text{ 1)}\\ {\text{(iv) 4yz(z}}^{\text{2}}\text{+ 6z }-\text{ 16) ÷ 2y(z + 8)}\\ {\text{(v) 5pq(p}}^{\text{2}}\text{ }-{\text{ q}}^{\text{2}}\text{) ÷ 2p(p + q)}\\ {\text{(vi) 12xy(9x}}^{\text{2}}\text{ }-{\text{ 16y}}^{\text{2}}\text{) ÷ 4xy(3x + 4y)}\\ {\text{(vii) 39y}}^{\text{3}}{\text{(50y}}^{\text{2}}-{\text{ 98) ÷ 26y}}^{\text{2}}\text{(5y + 7)}\end{array}$

Ans

$\begin{array}{l}\left(\mathrm{i}\right)\left({\mathrm{y}}^{2}+7\mathrm{y}+10\right)÷\left(\mathrm{y}+5\right)\\ \text{First we will factorise}\left({\mathrm{y}}^{2}+7\mathrm{y}+10\right)\\ ={\mathrm{y}}^{2}+2\mathrm{y}+5\mathrm{y}+10\\ =\mathrm{y}\left(\mathrm{y}+2\right)+5\left(\mathrm{y}+2\right)\\ =\left(\mathrm{y}+2\right)\left(\mathrm{y}+5\right)\\ \\ \therefore \frac{\left({\mathrm{y}}^{2}+7\mathrm{y}+10\right)}{\left(\mathrm{y}+5\right)}\\ =\frac{\left(\mathrm{y}+2\right)\overline{)\left(\mathrm{y}+5\right)}}{\overline{)\left(\mathrm{y}+5\right)}}\\ =\left(\mathrm{y}+2\right)\end{array}$

$\begin{array}{l}\left(\mathrm{ii}\right)\text{}\left({\mathrm{m}}^{2}-14\mathrm{m}-32\right)÷\left(\mathrm{m}+2\right)\\ \mathrm{First}\text{}\mathrm{we}\text{}\mathrm{will}\text{}\mathrm{factorise}\text{}\left({\mathrm{m}}^{2}-14\mathrm{m}-32\right)\\ ={\mathrm{m}}^{2}-16\mathrm{m}+2\mathrm{m}-32\\ =\mathrm{m}\left(\mathrm{m}-16\right)+2\left(\mathrm{m}-16\right)\\ =\left(\mathrm{m}-16\right)\left(\mathrm{m}+2\right)\\ \\ \therefore \frac{\left({\mathrm{m}}^{2}-14\mathrm{m}-32\right)}{\left(\mathrm{m}+2\right)}\\ =\frac{\left(\mathrm{m}-16\right)\overline{)\left(\mathrm{m}+2\right)}}{\overline{)\left(\mathrm{m}+2\right)}}\\ =\left(\mathrm{m}-16\right)\end{array}$ $\begin{array}{l}\end{array}$

$\begin{array}{l}\left(\mathrm{iii}\right)\text{}\left(5{\mathrm{p}}^{2}-25\mathrm{p}+20\right)÷\left(\mathrm{p}-1\right)\\ \mathrm{First}\text{}\mathrm{we}\text{}\mathrm{will}\text{}\mathrm{factorise}\text{}\left(5{\mathrm{p}}^{2}-25\mathrm{p}+20\right)\\ =5{\mathrm{p}}^{2}-20\mathrm{p}-5\mathrm{p}+20\\ =5\mathrm{p}\left(\mathrm{p}-4\right)-5\left(\mathrm{p}-4\right)\\ =5\left(\mathrm{p}-1\right)\left(\mathrm{p}-4\right)\\ \\ \therefore \frac{\left(5{\mathrm{p}}^{2}-25\mathrm{p}+20\right)}{\left(\mathrm{p}-1\right)}\\ =\frac{5\left(\mathrm{p}-4\right)\overline{)\left(\mathrm{p}-1\right)}}{\overline{)\left(\mathrm{p}-1\right)}}\\ =5\left(\mathrm{p}-4\right)\end{array}$ $\begin{array}{l}\end{array}$

$\begin{array}{l}\left(\mathrm{iv}\right)\text{}4\mathrm{yz}\left({\mathrm{z}}^{2}+6\mathrm{z}-16\right)÷2\mathrm{y}\left(\mathrm{z}+8\right)\\ \mathrm{First}\text{}\mathrm{we}\text{}\mathrm{will}\text{}\mathrm{factorise}\text{}4\mathrm{yz}\left({\mathrm{z}}^{2}+6\mathrm{z}-16\right)\\ =4\mathrm{yz}\left({\mathrm{z}}^{2}+8\mathrm{z}-2\mathrm{z}-16\right)\\ =4\mathrm{yz}\left[\mathrm{z}\left(\mathrm{z}+8\right)-2\left(\mathrm{z}+8\right)\right]\\ =4\mathrm{yz}\left[\left(\mathrm{z}+8\right)\left(\mathrm{z}-2\right)\right]\\ \\ \therefore \frac{4\mathrm{yz}\left({\mathrm{z}}^{2}+6\mathrm{z}-16\right)}{2\mathrm{y}\left(\mathrm{z}+8\right)}\\ =\frac{4\overline{)\mathrm{y}}\mathrm{z}\left[\overline{)\left(\mathrm{z}+8\right)}\left(\mathrm{z}-2\right)\right]}{2\overline{)\mathrm{y}}\overline{)\left(\mathrm{z}+8\right)}}\\ =2\mathrm{z}\left(\mathrm{z}-2\right)\end{array}$ $\begin{array}{l}\end{array}$

$\begin{array}{l}\left(\mathrm{v}\right)\text{}5\mathrm{pq}\left({\mathrm{p}}^{2}-{\mathrm{q}}^{2}\right)÷2\mathrm{p}\left(\mathrm{p}+\mathrm{q}\right)\\ \mathrm{First}\text{}\mathrm{we}\text{}\mathrm{will}\text{}\mathrm{factorise}\text{}5\mathrm{pq}\left({\mathrm{p}}^{2}-{\mathrm{q}}^{2}\right)\\ =5\mathrm{pq}\left(\mathrm{p}-\mathrm{q}\right)\left(\mathrm{p}+\mathrm{q}\right)\text{\hspace{0.17em}}\left[\mathrm{Byusing}:{\mathrm{a}}^{2}-{\mathrm{b}}^{2}=\left(\mathrm{a}+\mathrm{b}\right)\left(\mathrm{a}-\mathrm{b}\right)\right]\\ \\ \therefore \frac{5\mathrm{pq}\left({\mathrm{p}}^{2}-{\mathrm{q}}^{2}\right)}{2\mathrm{p}\left(\mathrm{p}+\mathrm{q}\right)}\\ =\frac{5\overline{)\mathrm{p}}\mathrm{q}\left(\mathrm{p}-\mathrm{q}\right)\overline{)\left(\mathrm{p}+\mathrm{q}\right)}\text{\hspace{0.17em}}}{2\overline{)\mathrm{p}}\overline{)\left(\mathrm{p}+\mathrm{q}\right)}}\\ =\frac{5\mathrm{q}\left(\mathrm{p}-\mathrm{q}\right)}{2}\\ \end{array}$ $\begin{array}{l}\end{array}$

$\begin{array}{l}\left(\mathrm{vi}\right)\text{}12\mathrm{xy}\left(9{\mathrm{x}}^{2}-16{\mathrm{y}}^{2}\right)÷4\mathrm{xy}\left(3\mathrm{x}+4\mathrm{y}\right)\\ \mathrm{First}\text{}\mathrm{we}\text{}\mathrm{will}\text{}\mathrm{factorise}\text{}12\mathrm{xy}\left(9{\mathrm{x}}^{2}-16{\mathrm{y}}^{2}\right)\\ =12\mathrm{xy}\left[{\left(3\mathrm{x}\right)}^{2}-{\left(4\mathrm{y}\right)}^{2}\right]\\ =\text{}12\mathrm{xy}\left[\left(3\mathrm{x}-4\mathrm{y}\right)\left(3\mathrm{x}+4\mathrm{y}\right)\right]\text{}\left[\mathrm{Byusing}:{\mathrm{a}}^{2}-{\mathrm{b}}^{2}=\left(\mathrm{a}+\mathrm{b}\right)\left(\mathrm{a}-\mathrm{b}\right)\right]\\ \\ \therefore \frac{12\overline{)\mathrm{xy}}\left[\overline{)\left(3\mathrm{x}+4\mathrm{y}\right)}\left(3\mathrm{x}-4\mathrm{y}\right)\right]\text{}}{4\overline{)\mathrm{xy}}\overline{)\left(3\mathrm{x}+4\mathrm{y}\right)}}\\ =3\left(3\mathrm{x}-4\mathrm{y}\right)\end{array}$ $\begin{array}{l}\end{array}$

$\begin{array}{l}\left(\mathrm{vii}\right)\text{}39{\mathrm{y}}^{3}\left(50{\mathrm{y}}^{2}-98\right)÷26{\mathrm{y}}^{2}\left(5\mathrm{y}+7\right)\\ \mathrm{First}\text{}\mathrm{we}\text{}\mathrm{will}\text{}\mathrm{factorise}\text{}39{\mathrm{y}}^{3}\left(50{\mathrm{y}}^{2}-98\right)\\ =13×3×{\mathrm{y}}^{3}×2\left(25{\mathrm{y}}^{2}-49\right)\\ =13×3×{\mathrm{y}}^{3}×2\left[{\left(5\mathrm{y}\right)}^{2}-{7}^{2}\right]\\ =13×3×{\mathrm{y}}^{3}×2\left[\left(5\mathrm{y}-7\right)\left(5\mathrm{y}+7\right)\right]\text{}\\ \text{\hspace{0.17em}}\left[\mathrm{Byusing}:{\mathrm{a}}^{2}-{\mathrm{b}}^{2}=\left(\mathrm{a}+\mathrm{b}\right)\left(\mathrm{a}-\mathrm{b}\right)\right]\\ \\ \therefore \frac{39{\mathrm{y}}^{3}\left(50{\mathrm{y}}^{2}-98\right)}{26{\mathrm{y}}^{2}\left(5\mathrm{y}+7\right)}\end{array}$ $\begin{array}{l}\end{array}$

$\begin{array}{l}=\frac{\overline{)13}×3×{\mathrm{y}}^{3}×\overline{)2}\left[\left(5\mathrm{y}-7\right)\overline{)\left(5\mathrm{y}+7\right)}\right]\text{}}{\overline{)2}×\overline{)13}×{\mathrm{y}}^{2}\overline{)\left(5\mathrm{y}+7\right)}}\\ =3\mathrm{y}\left(5\mathrm{y}-7\right)\\ \\ \end{array}$

FAQs (Frequently Asked Questions)

1. Are the NCERT Solutions For Class 8 Mathematics Chapter 14 Exercise 14.3 important from an exam perspective?

Yes, factorization is important from an exam perspective. Long answer questions are included as well as short answer questions to improve problem solving skills. These solutions are created by Extramarks’s expert faculty to help students prepare for their exams.

2. How many exercises are there in NCERT Solutions for Class 8 Mathematics Chapter 14?

NCERT Mathematics Class 8 Chapter 14 Answers has 4 exercises, these answers will help students to solve problems skill-fully and efficiently. It also focuses on solving math solutions in a way that is easy for students to understand.

3. What are the main topics covered in Chapter 14 Mathematics?

The main topics covered in Chapter 14 Mathematics are:

14.1 – Introduction

14.2 – Definition of factorization

14.3 – Algebraic division

14.4 – Algebraic division