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# NCERT Solutions for Class 8 Maths Chapter 12 Factorisation

Last updated date: 19th Jun 2024
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## Complete Resource of NCERT Solutions Maths Chapter 12 Factorisation Class 8 - FREE PDF Download

NCERT Solutions is to help students capture the material while developing a careful understanding of the different kinds of questions that are asked in the CBSE Factorisation Class 8 Mathematics Examinations. There are suitable answers for every question, which will be helpful when updating the CBSE Class 8 Mathematics syllabus. According to the CBSE instruction, highly qualified subject matter experts create the Math Class 8 solutions. Students can simplify their final-minute review by using the Class 8 Maths NCERT Solutions. Students who wish to study offline can download the PDF.

Table of Content
1. Complete Resource of NCERT Solutions Maths Chapter 12 Factorisation Class 8 - FREE PDF Download
2. Glance of NCERT Solutions for Class 8 Maths Chapter 12 Factorisation
3. Access Exercise Wise NCERT Solutions for Chapter 12 Maths Class 8
4. Exercises Under NCERT Solutions for Class 8 Maths Chapter 12 Factoristaion
5. Access NCERT Solutions for Class 8 Maths Chapter 12 – Factorisation
5.1Exercise 12.1
5.2Exercise 12.2
6. Overview of Deleted Syllabus for CBSE Factorisation Class 8 Maths Factorisation
7. Class 8 Maths Chapter 12: Exercises Breakdown
8. Other Study Material for CBSE Class 8 Maths Chapter 12
9. Chapter-Specific NCERT Solutions for Class 8 Maths
FAQs

## Glance of NCERT Solutions for Class 8 Maths Chapter 12 Factorisation

1. The formulas used in this chapter are:

• Common Factor for identifying the greatest common divisor (GCD) of terms.

• Factorisation of Quadratic Expressions that is

$ax^{^{2}}+bx+c= \left ( px+q \right )\left ( rx+s \right )$

• Factorisation by Grouping terms to factorise polynomials.

1. This chapter focuses on finding Common Factors like simplifying expressions by extracting common factors.

2. Focuses on basic factorisation techniques, essential for building foundational skills.

3. It deals with the factorisation of quadratic expressions, crucial for understanding advanced algebra.

4. This chapter covers advanced factorisation methods, including grouping, enhancing problem-solving skills.

5. This chapter applies factorisation in polynomial division, important for comprehensive understanding of algebraic operations.

6. There are Exercises links provided. It has solutions for each question from Factorisation.

7. There are three exercises (13 fully solved questions) in Chapter 12 Factorisation Class 8 Maths.

## Access Exercise Wise NCERT Solutions for Chapter 12 Maths Class 8

 S.No. Current Syllabus Exercises of Class 8 Maths Chapter 12 1 NCERT Solutions of Class 8 Maths Chapter 12 Factorisation Exercise 12.1 2 NCERT Solutions of Class 8 Maths Chapter 12 Factorisation Exercise 12.2 3 NCERT Solutions of Class 8 Maths Chapter 12 Factorisation Exercise 12.3
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## Exercises Under NCERT Solutions for Class 8 Maths Chapter 12 Factoristaion

### Introduction

The first section of every chapter might be the same in most cases. It is the Introduction part. Here the introduction may remind the knowledge of factors for the students which they have studied in the earlier classes. Students have already learned to find factors for natural numbers and the meaning of factors for algebraic expressions. Then in this chapter, they are going to learn the factorisation of big expressions using different methods in detail.

### 12.1 What is Factorisation?

The Class 8 Maths Chapter Factorisation Pdf explains the meaning of factorisation. The factorisation is a process of representing the given expression in the form of the product of two factors. These factors may be either of the numbers or variables or algebraic expressions themselves. The PDF also explains that it has several systematic approaches to find factors for the given expressions. Let's see the available methods to find the factors.

### 12.1.1 Method of Common Factors

It is the first method of finding factors for the given expression. According to NCERT Solutions Class 8 Maths Chapter Factorisation , the given expression can be expressed in the product of terms. This can be understood clearly by observing two or more examples given in the PDF.

### 12.2 Factorization by Regrouping Terms

The next section of this chapter deals with another method of factorization. Chapter 12 Class 8 Maths Factorisation explains that while finding the factors of the given expression, we will take an individual term as common and consider it as a common factor. This can be done by forming two groups. For an instance,

6𝑝 − 12𝑞

= 6𝑝 − (6 × 2)𝑞

= 6(𝑝 − 2𝑞)(taking 6 as common)

Exercise 14.1 with three short answer type questions.

### 12.2.1 Factorization Using Identities

NCERT Class 8 Maths Factorisation Chapter 12 came to the third method of factorization in this section. Here students need to find out the identical terms and then separate those terms which give the factors directly.

### 12.2.2 Factors of the form (x+a)(x+b)

This is the 4th method of solving factorization. The Class 8 Maths Factorisation Chapter 12 explains that this method was usually used if the expression doesn't have perfect squares. Even though it is suitable for perfect squares, without having perfect squares, also we can use this method to get factors of the given expression. Even though we have different methods to find factors, the solution is one and the same, but the procedures may differ.

Exercise 14.2 has five short answer type questions in which each question has some set of expressions to solve.

### 12.3 Division of Algebraic Expressions

The NCERT Solutions For Class 8 Maths Factorisation Chapter 12 explains an advanced concept in factorization. As the students have learned addition, subtraction, multiplication of algebraic expressions, now class 8 maths factorization explains how to divide two or more expressions. But it has two sub-concepts. Because the division can be made based on the degree of expression. From the notes of NCERT Solutions Chapter 12 Class 8, we have:

• Division of monomial by another monomial.

• Division of polynomial by monomial.

### 12.3.1 Division of Algebraic Expressions Continued

Class 8 Maths Chapter Factorisation Solutions has explained that this section is the extension of the above section. Here the division of expressions can be made by two or more polynomials. It explains how to divide the Algebraic Expressions of higher degrees.

## Access NCERT Solutions for Class 8 Maths Chapter 12 – Factorisation

### Exercise 12.1

1. Find the common factors of the terms

i. $12x,36$

Ans:  Write the factors of each term separately:

$12x=2\times 2\times 3\times x$

$36=2\times 2\times 3\times 3$

The factors that appear in both the lists are the common factors.

Hence, the common factors are $2,2,3$.

Multiply the common factors, $2\times 2\times 3=12$

ii.  $2y,22xy$

Ans: Write the factors of each term separately:

$2y=2\times y$

$22xy=2\times 11\times x\times y$

The factors that appear in both the lists are the common factors.

Hence, the common factors are $2,y$.

Multiply the common factors, $2\times y=2y$.

iii. $14pq,28{{p}^{2}}{{q}^{2}}$

Ans: Write the factors of each term separately:

$14pq=2\times 7\times p\times q$

$28{{p}^{2}}{{q}^{2}}=2\times 2\times 7\times p\times p\times q\times q$

The factors that appear in both the lists are the common factors.

Hence, the common factors are$2,7,p,q$.

Multiply the common factors, $2\times 7\times p\times q=14pq$.

iv. $2x,3{{x}^{2}},4$

Ans: Write the factors of each term separately:

$2x=2\times x$

$3{{x}^{2}}=3\times x\times x$

$4=2\times 2$

The factors that appear in both the lists are the common factors.

Hence, the common factor is $1$.

v. $6abc,24a{{b}^{2}},12{{a}^{2}}b$

Ans: Write the factors of each term separately:

$6abc=2\times 3\times a\times b\times c$

$24a{{b}^{2}}=2\times 2\times 2\times 3\times a\times b\times b$

$12{{a}^{2}}b=2\times 2\times 3\times a\times a\times b$

The factors that appear in both the lists are the common factors.

Hence, the common factors are$2,3,a,b$.

Multiply the common factors, $2\times 3\times a\times b=6ab$.

vi. $16{{x}^{3}},-4{{x}^{2}},32x$

Ans: Write the factors of each term separately:

$16{{x}^{3}}=2\times 2\times 2\times 2\times x\times x\times x$

$-4{{x}^{2}}=-1\times 2\times 2\times x\times x$

$32x=2\times 2\times 2\times 2\times 2\times x$

The factors that appear in both the lists are the common factors.

Hence, the common factors are$2,2,x$

Multiply the common factors, $2\times 2\times x=4x$

vii. $10pq,20qr,30rp$

Ans: Write the factors of each term separately:

$10pq=2\times 5\times p\times q$

$20qr=2\times 2\times 5\times q\times r$

$30rp=2\times 3\times 5\times r\times p$

The factors that appear in both the lists are the common factors.

Hence, the common factors are $2,5$.

Multiply the common factors, $2\times 5=10$.

viii. $3{{x}^{2}}{{y}^{3}},10{{x}^{3}}{{y}^{2}},6{{x}^{2}}{{y}^{2}}z$

Ans: Write the factors of each term separately:

$3{{x}^{2}}{{y}^{3}}=3\times x\times x\times y\times y\times y$

$10{{x}^{3}}{{y}^{2}}=2\times 5\times x\times x\times x\times y\times y$

$6{{x}^{2}}{{y}^{2}}z=2\times 3\times x\times x\times y\times y\times z$

The factors that appear in both the lists are the common factors.

Hence, the common factors are $2\times 7\times p\times q=14pq$.

2. Factorise the following expressions

i. $7x-42$

Ans: Take out the common factors from all the terms to factorize.

$\text{ }7x=7\times x$

$42=2\times 3\times 7$

The common factor is 7

$\therefore 7x-42=(7\times x)-(2\times 3\times 7)=7(x-6)$

ii. $6p-12q$

Ans: Take out the common factors from all the terms to factorize.

$6p=2\times 3\times p$

$12q=2\times 2\times 3\times q$

The common factors are $2$ and $3$.

$6p-12q=(2\times 3\times p)-(2\times 2\times 3\times q)$

$=2\times 3[p-(2\times q)]$

$=6(p-2q)$

iii. $7{{a}^{2}}+14a$

Ans: Take out the common factors from all the terms to factorize.

$7{{a}^{2}}=7\times a\times a$

$14a=2\times 7\times a$

The common factors are $7$ and $a$

$\therefore 7{{a}^{2}}+14a=(7\times a\times a)+(2\times 7\times a)$

$=7\times a[a+2]$

$=7a(a+2)$

iv. $-16z+20{{z}^{3}}$

Ans: Take out the common factors from all the terms to factorize.

$\text{ }16z=2\times 2\times 2\times 2\times z$

$20{{z}^{3}}=2\times 2\times 5\times z\times z\times z$

The common factors are $2,2$, and $z$ .

$-16z+20{{z}^{3}}=-(2\times 2\times 2\times 2\times z)+(2\times 2\times 5\times z\times z\times z)$

$=(2\times 2\times z)[-(2\times 2)+(5\times z\times z)]$

$=4z\left( -4+5{{z}^{2}} \right)$

v. $20{{l}^{2}}m+30\text{alm}$

Ans: Take out the common factors from all the terms to factorize.

$20{{l}^{2}}m=2\times 2\times 5\times 1\times 1\times m\text{ }$

$\text{30alm }=2\times 3\times 5\times a\times 1\times m$

The common factors are $2,5,1,$ and $m$.

$\therefore 20{{l}^{2}}m+30alm=(2\times 2\times 5\times 1\times 1\times m)+(2\times 3\times 5\times a\times 1\times m)$

$=(2\times 5\times 1\times m)[(2\times l)+(3\times a)]$

$=10lm(2l+3a)$

vi. $5{{x}^{2}}y-15x{{y}^{2}}$

Ans: Take out the common factors from all the terms to factorize.

$5{{x}^{2}}y=5\times x\times x\times y$

$15x{{y}^{2}}=3\times 5\times x\times y\times y$

The common factors are$5,x$, and$y$.

$5{{x}^{2}}y-15x{{y}^{2}}=(5\times x\times x\times y)-(3\times 5\times x\times y\times y)$

$=5\times x\times y[x-(3\times y)]$

$=5xy(x-3y)$

vii. $10{{a}^{2}}-15{{b}^{2}}+20{{c}^{2}}$

Ans: Take out the common factors from all the terms to factorize.

$10{{a}^{2}}=2\times 5\times a\times a$

$15{{b}^{2}}=3\times 5\times b\times b$

$20{{c}^{2}}=2\times 2\times 5\times c\times c$

The common factor is $5$

$10{{a}^{2}}-15{{b}^{2}}+20{{c}^{2}}=(2\times 5\times a\times a)-(3\times 5\times b\times b)+(2\times 2\times 5\times c\times c)$

$=5[(2\times a\times a)-(3\times b\times b)+(2\times 2\times c\times c)]$

$=5\left( 2{{a}^{2}}-3{{b}^{2}}+4{{c}^{2}} \right)$

viii. $-4{{a}^{2}}+4ab-4ca$

Ans: Take out the common factors from all the terms to factorize.

$4{{a}^{2}}=2\times 2\times a\times a$

$4ab=2\times 2\times a\times b$

$4ca=2\times 2\times c\times a$

The common factors are $2,2$, and $a$

$-4{{a}^{2}}+4ab-4ca=-(2\times 2\times a\times a)+(2\times 2\times a\times b)-(2\times 2\times c\times a)$

$=2\times 2\times a[-(a)+b-c]$

$=4a(-a+b-c)\text{ }$

ix. ${{x}^{2}}yz+x{{y}^{2}}z+xy{{z}^{2}}$

Ans: Take out the common factors from all the terms to factorize.

${{x}^{2}}yz=x\times x\times y\times z$

$x{{y}^{2}}z=x\times y\times y\times z$

$xy{{z}^{2}}=x\times y\times z\times z$

The common factors are$x,y$, and $z$

${{x}^{2}}yz+x{{y}^{2}}z+xy{{z}^{2}}=(x\times x\times y\times z)+(x\times y\times y\times z)+(x\times y\times z\times z)$

$=x\times y\times z[x+y+z]$

$=xyz(x+y+z)$

x. $a{{x}^{2}}y+bx{{y}^{2}}+cxyz$

Ans: Take out the common factors from all the terms to factorize.

$a{{x}^{2}}y=a\times x\times x\times y$

$bx{{y}^{2}}=b\times x\times y\times y$

$\operatorname{cxyz}=c\times x\times y\times z$

The common factors are$x$and $y$.

$a{{x}^{2}}y+bx{{y}^{2}}+cxyz=(a\times x\times x\times y)+(b\times x\times y\times y)+(c\times x\times y\times z)$

$=(x\times y)[(a\times x)+(b\times y)+(c\times z)]$

$=xy(ax+by+cz)$

3.  Factorise

i. ${{x}^{2}}+xy+8x+8y$

Ans:

Write each term in terms of its factors and take out the common factors.

${{x}^{2}}+xy+8x+8y=x\times x+x\times y+8\times x+8\times y$

$=x(x+y)+8(x+y)$

$=(x+y)(x+8)$

ii. $15xy-6x+5y-2$

Ans:

Write each term in terms of its factors and take out the common factors.

$15xy-6x+5y-2=3\times 5\times x\times y-3\times 2\times x+5\times y-2$

$=3x(5y-2)+1(5y-2)$

$=(5y-2)(3x+1)$

iii. $ax+bx-ay-by$

Ans:

Write each term in terms of its factors and take out the common factors.

$ax+bx-ay-by=a\times x+b\times x-a\times y-b\times y$

$=x(a+b)-y(a+b)$

$=(a+b)(x-y)$

iv. $15pq+15+9q+25p$

Ans:

Write each term in terms of its factors and take out the common factors.

$15pq+15+9q+25p=15pq+9q+25p+15$

$=3\times 5\times p\times q+3\times 3\times q+5\times 5\times p+3\times 5$

$=3\text{ }q\left( 5\text{ }p+3 \right)+5\left( 5\text{ }p+3 \right)$

$=\left( 5\text{ }p+3 \right)\left( 3\text{ }q+5 \right)$

v.  $z-7+7xy-xyz$

Ans:

Write each term in terms of its factors and take out the common factors.

$z-7+7xy-xyz=z-x\times y\times z-7+7\times x\times y$

$=z\left( 1-x\text{ }y \right)-7\left( 1-x\text{ }y \right)$

$=(1-xy)(z-7)$

### Exercise 12.2

1. Factorise the following expressions.

i. ${{a}^{2}}+8a+16$

Ans: Use the formula $\left[ {{(x+y)}^{2}}={{x}^{2}}+2xy+{{y}^{2}} \right]$ to factorize.

${{a}^{2}}+8a+16={{(a)}^{2}}+2\times a\times 4+{{(4)}^{2}}$

$={{(a+4)}^{2}}$

ii. ${{p}^{2}}-10p+25$

Ans: Use the formula $\left[ {{(x+y)}^{2}}={{x}^{2}}+2xy+{{y}^{2}} \right]$ to factorize.

${{p}^{2}}-10p+25={{(p)}^{2}}-2\times p\times 5+{{(5)}^{2}}$

$={{(p-5)}^{2}}$

iii. $25{{m}^{2}}+30m+9$

Ans: Use the formula $\left[ {{(x+y)}^{2}}={{x}^{2}}+2xy+{{y}^{2}} \right]$ to factorize.

$25{{m}^{2}}+30m+9={{(5m)}^{2}}+2\times 5m\times 3+{{(3)}^{2}}$

$={{(5m+3)}^{2}}$

iv. $49{{y}^{2}}+84yz+36{{z}^{2}}$

Ans: Use the formula $\left[ {{(x+y)}^{2}}={{x}^{2}}+2xy+{{y}^{2}} \right]$ to factorize.

$49{{y}^{2}}+84yz+36{{z}^{2}}={{(7y)}^{2}}+2\times (7y)\times (6z)+{{(6z)}^{2}}$

$={{(11b-4c)}^{2}}$

v. $4{{x}^{2}}-8x+4$

Ans: Use the formula $\left[ {{(x+y)}^{2}}={{x}^{2}}+2xy+{{y}^{2}} \right]$ to factorize.

$4{{x}^{2}}-8x+4={{(2x)}^{2}}-2(2x)(2)+{{(2)}^{2}}$

$={{(2x-2)}^{2}}$

$={{[(2)(x-1)]}^{2}}$

$=4{{(x-1)}^{2}}$

vi. $121{{b}^{2}}-88bc+16{{c}^{2}}$

Ans: Use the formula $\left[ {{(x+y)}^{2}}={{x}^{2}}+2xy+{{y}^{2}} \right]$ to factorize.

$121{{b}^{2}}-88bc+16{{c}^{2}}={{(11b)}^{2}}-2(11b)(4c)+{{(4c)}^{2}}$

$={{(11b-4c)}^{2}}$

vii. ${{(l+m)}^{2}}-4lm$ (Hint: Expand ${{(l+m)}^{2}}$first)

Ans: Use identity $\left[ {{(a+b)}^{2}}={{a}^{2}}+2ab+{{b}^{2}} \right]$to factorize.

${{(l+m)}^{2}}-4lm={{l}^{2}}+2lm+{{m}^{2}}-4lm$

$={{l}^{2}}-2lm+{{m}^{2}}$

$={{(l-m)}^{2}}$

viii. ${{a}^{4}}+2{{a}^{2}}{{b}^{2}}+{{b}^{4}}$

Ans: Use identity $\left[ {{(a+b)}^{2}}={{a}^{2}}+2ab+{{b}^{2}} \right]$to factorize.

${{a}^{4}}+2{{a}^{2}}{{b}^{2}}+{{b}^{4}}={{\left( {{a}^{2}} \right)}^{2}}+2\left( {{a}^{2}} \right)\left( {{b}^{2}} \right)+{{\left( {{b}^{2}} \right)}^{2}}$

$={{\left( {{a}^{2}}+{{b}^{2}} \right)}^{2}}$

2. Factorise

i. $4{{p}^{2}}-9{{q}^{2}}$

Ans: Use the identity $\left[ {{a}^{2}}-{{b}^{2}}=(a-b)(a+b) \right]$ for factorizing.

$4{{p}^{2}}-9{{q}^{2}}={{(2p)}^{2}}-{{(3q)}^{2}}$

$=(2p+3q)(2p-3q)$

ii. $63{{a}^{2}}-112{{b}^{2}}$

Ans: Use the identity $\left[ {{a}^{2}}-{{b}^{2}}=(a-b)(a+b) \right]$ for factorizing.

$63{{a}^{2}}-112{{b}^{2}}=7\left( 9{{a}^{2}}-16{{b}^{2}} \right)$

$=7\left[ {{(3a)}^{2}}-{{(4b)}^{2}} \right]$

$=7(3a+4b)(3a-4b)$

iii. $49{{x}^{2}}-36$

Ans: Use the identity $\left[ {{a}^{2}}-{{b}^{2}}=(a-b)(a+b) \right]$ for factorizing.

$49{{x}^{2}}-36={{(7x)}^{2}}-{{(6)}^{2}}$

$=(7x-6)(7x+6)$

iv. $16{{x}^{5}}-144{{x}^{3}}$

Ans: Use the identity $\left[ {{a}^{2}}-{{b}^{2}}=(a-b)(a+b) \right]$ for factorizing.

$16{{x}^{5}}-144{{x}^{3}}=16{{x}^{3}}\left( {{x}^{2}}-9 \right)$

$=16{{x}^{3}}\left[ {{(x)}^{2}}-{{(3)}^{2}} \right]$

$=16{{x}^{3}}(x-3)(x+3)$

v. ${{(l+m)}^{2}}-{{(l-m)}^{2}}$

Ans: Use the identity $\left[ {{a}^{2}}-{{b}^{2}}=(a-b)(a+b) \right]$ for factorizing.

${{(l+m)}^{2}}-{{(l-m)}^{2}}=[(l+m)-(l-m)][(l+m)+(l-m)]$ $=\left( l+m-1+m \right)\left( I+m+l-m \right)$

$=2m\times 2l$

$=4\text{ml}$

$=4lm$

vi. $9{{x}^{2}}{{y}^{2}}-16$

Ans: Use the identity $\left[ {{a}^{2}}-{{b}^{2}}=(a-b)(a+b) \right]$ for factorizing.

$9{{x}^{2}}{{y}^{2}}-16={{(3xy)}^{2}}-{{(4)}^{2}}$

$=(3xy-4)(3xy+4)$

vii. $\left( {{x}^{2}}-2xy+{{y}^{2}} \right)-{{z}^{2}}$

Ans: Use the identity $\left[ {{a}^{2}}-{{b}^{2}}=(a-b)(a+b) \right]$ for factorizing.

$\left( {{x}^{2}}-2xy+{{y}^{2}} \right)-{{z}^{2}}={{(x-y)}^{2}}-{{(z)}^{2}}$

$=(x-y-z)(x-y+z)$

viii. $25{{a}^{2}}-4{{b}^{2}}+28bc-49{{c}^{2}}$

Ans: Use the identity $\left[ {{a}^{2}}-{{b}^{2}}=(a-b)(a+b) \right]$ for factorizing.

$25{{a}^{2}}-4{{b}^{2}}+28bc-49{{c}^{2}}=25{{a}^{2}}-\left( 4{{b}^{2}}-28bc+49{{c}^{2}} \right)$   $={{(5a)}^{2}}-\left[ {{(2b)}^{2}}-2\times 2b\times 7c+{{(7c)}^{2}} \right]$

$={{(5a)}^{2}}-\left[ {{(2b-7c)}^{2}} \right]$

$=\left[ 5\text{ }a+\left( 2\text{ }b-7\text{ }c \right) \right]\left[ 5\text{ }a-\left( 2\text{ }b-7\text{ }c \right) \right]$

$=\left( 5\text{ }a+2\text{ }b-7\text{ }c \right)\left( 5\text{ }a-2\text{ }b+7\text{ }c \right)$

3. Factorise the expressions

i. $a{{x}^{2}}+bx$

Ans: Take the common factors out to factorize.

$a{{x}^{2}}+bx=a\times x\times x+b\times x$

$=x(ax+b)$

ii. $7{{p}^{2}}+21{{q}^{2}}$

Ans: Take the common factors out to factorize.

$7{{p}^{2}}+21{{q}^{2}}=7\times p\times p+3\times 7\times q\times q$

$=7\left( {{p}^{2}}+3{{q}^{2}} \right)$

iii. $2{{x}^{3}}+2x{{y}^{2}}+2x{{z}^{2}}$

Ans: Take the common factors out to factorize.

$2{{x}^{3}}+2x{{y}^{2}}+2x{{z}^{2}}=2x\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)$

iv. $a{{m}^{2}}+b{{m}^{2}}+b{{n}^{2}}+a{{n}^{2}}$

Ans: Take the common factors out to factorize.

$a{{m}^{2}}+b{{m}^{2}}+b{{n}^{2}}+a{{n}^{2}}=a{{m}^{2}}+b{{m}^{2}}+a{{n}^{2}}+b{{n}^{2}}$

$={{m}^{2}}(a+b)+{{n}^{2}}(a+b)$

$=(a+b)\left( {{m}^{2}}+{{n}^{2}} \right)$

v. $(lm+l)+m+l$

Ans: Take the common factors out to factorize.

$(lm+l)+m+1=1m+m+1+1$

$=m(l+1)+1(l+1)$

$=(l+1)(m+1)$

vi. $y(y+z)+9(y+z)$

Ans: Take the common factors out to factorize.

$y(y+z)+9(y+z)=(y+z)(y+9)$

vii. $5{{y}^{2}}-20y-8z+2yz$

Ans: Take the common factors out to factorize.

$5{{y}^{2}}-20y-8z+2yz=5{{y}^{2}}-20y+2yz-8z$

$=5y(y-4)+2z(y-4)$

$=(y-4)(5y+2z)$

viii. $10ab+4a+5b+2$

Ans: Take the common factors out to factorize.

$10ab+4a+5b+2=10ab+5b+4a+2$

$=5\text{ }b\left( 2\text{ }a+1 \right)+2\left( 2\text{ }a+1 \right)$

$=\left( 2\text{ }a+1 \right)\left( 5\text{ }b+2 \right)$

ix. $6xy-4y+6-9x$

Ans: Take the common factors out to factorize.

$6xy-4\text{ }y+6-9x=6xy-9x-4y+6$

$=3x\left( 2y-3 \right)-2\left( 2y-3 \right)$

$=\left( 2y-3 \right)\left( 3x-2 \right)$

4. Factorise

i. ${{a}^{4}}-{{b}^{4}}$

Ans: Use the identity $\left[ {{a}^{2}}-{{b}^{2}}=(a-b)(a+b) \right]$ for factorizing.

${{a}^{4}}-{{b}^{4}}={{\left( {{a}^{2}} \right)}^{2}}-{{\left( {{b}^{2}} \right)}^{2}}$

$=\left( {{a}^{2}}-{{b}^{2}} \right)\left( {{a}^{2}}+{{b}^{2}} \right)$

$=(a-b)(a+b)\left( {{a}^{2}}+{{b}^{2}} \right)$

ii. ${{p}^{4}}-81$

Ans: Use the identity $\left[ {{a}^{2}}-{{b}^{2}}=(a-b)(a+b) \right]$ for factorizing.

${{p}^{4}}-81={{\left( {{p}^{2}} \right)}^{2}}-{{(9)}^{2}}$

$=\left( {{p}^{2}}-9 \right)\left( {{p}^{2}}+9 \right)$

$=\left[ {{(p)}^{2}}-{{(3)}^{2}} \right]\left( {{p}^{2}}+9 \right)$

$=(p-3)(p+3)\left( {{p}^{2}}+9 \right)$

iii. ${{x}^{4}}-{{(y+z)}^{4}}$

Ans: Use the identity $\left[ {{a}^{2}}-{{b}^{2}}=(a-b)(a+b) \right]$ for factorizing.

${{x}^{4}}-{{(y+z)}^{4}}={{\left( {{x}^{2}} \right)}^{2}}-{{\left[ {{(y+z)}^{2}} \right]}^{2}}$

$=\left[ {{x}^{2}}-{{(y+z)}^{2}} \right]\left[ {{x}^{2}}+{{(y+z)}^{2}} \right]$

$=[x-(y+z)][x+(y+z)]\left[ {{x}^{2}}+{{(y+z)}^{2}} \right]$

$=(x-y-z)(x+y+z)\left[ {{x}^{2}}+{{(y+z)}^{2}} \right]$

iv. ${{x}^{4}}-{{(x-z)}^{4}}$

Ans: Use the identity $\left[ {{a}^{2}}-{{b}^{2}}=(a-b)(a+b) \right]$ for factorizing.

${{x}^{4}}-{{(x-z)}^{4}}={{\left( {{x}^{2}} \right)}^{2}}-{{\left[ {{(x-z)}^{2}} \right]}^{2}}$

$=\left[ {{x}^{2}}-{{(x-z)}^{2}} \right]\left[ {{x}^{2}}+{{(x-z)}^{2}} \right]$

$=[x-(x-z)][x+(x-z)]\left[ {{x}^{2}}+{{(x-z)}^{2}} \right]$

$=z(2x-z)\left[ {{x}^{2}}+{{x}^{2}}-2xz+{{z}^{2}} \right]$

$=z(2x-z)\left( 2{{x}^{2}}-2xz+{{z}^{2}} \right)$

v. ${{a}^{4}}-2{{a}^{2}}{{b}^{2}}+{{b}^{4}}$

Ans: Use the identity $\left[ {{a}^{2}}-{{b}^{2}}=(a-b)(a+b) \right]$ for factorizing.${{a}^{4}}-2{{a}^{2}}{{b}^{2}}+{{b}^{4}}={{\left( {{a}^{2}} \right)}^{2}}-2\left( {{a}^{2}} \right)\left( {{b}^{2}} \right)+{{\left( {{b}^{2}} \right)}^{2}}$

$={{\left( {{a}^{2}}-{{b}^{2}} \right)}^{2}}$

$={{[(a-b)(a+b)]}^{2}}$

$={{(a-b)}^{2}}{{(a+b)}^{2}}$

5. Factorise the following expressions

i. ${{p}^{2}}+6p+8$

Ans: We can see that, $8=4\times 2$and $4+2=6$

$\therefore {{p}^{2}}+6p+8={{p}^{2}}+2p+4p+8$

$=p\left( p+2 \right)+4\left( p+2 \right)$

$=\left( p+2 \right)\left( p+4 \right)$

ii. ${{q}^{2}}-10q+21$

Ans: We can see that, $21=(-7)\times (-3)$and $(-7)+(-3)=-10$

$\therefore {{q}^{2}}-10q+21={{q}^{2}}-7q-3q+21$

$=q\left( q-7 \right)-3\left( q-7 \right)$

$=\left( q-7 \right)\left( q-3 \right)$

iii. ${{p}^{2}}+6p-16$

Ans: We can see that, $16=(-2)\times 8$ and $8+(-2)=6$

${{p}^{2}}+6p-16={{p}^{2}}+8p-2p-16$

$=p\left( p+8 \right)-2\left( p+8 \right)$

$=\left( p+8 \right)\left( p-2 \right)$

Exercise (12.3)

1. Carry out the following divisions.

i. $28{{x}^{4}}\div 56x$

Ans: Write the numerator and denominator in its factors and divide.

$28{{x}^{4}}=2\times 2\times 7\times x\times x\times x\times x$

$56x=2\times 2\times 2\times 7\times x$

$28{{x}^{4}}\div 56x=\dfrac{2\times 2\times 7\times x\times x\times x\times x}{2\times 2\times 2\times 7\times x}$

$=\dfrac{{{x}^{3}}}{2}$

$=\dfrac{1}{2}{{x}^{3}}$

ii. $-36{{y}^{3}}\div 9{{y}^{2}}$

Ans: Write the numerator and denominator in its factors and divide.

$36{{y}^{3}}=2\times 2\times 3\times 3\times y\times y\times y$

$9{{y}^{2}}=3\times 3\times y\times y$

$-36{{y}^{3}}\div 9{{y}^{2}}=\dfrac{-2\times 2\times 3\times 3\times y\times y\times y}{3\times 3\times y\times y}$

$=-4y$

iii. $66p{{q}^{2}}{{r}^{3}}\div 11q{{r}^{2}}$

Ans: Write the numerator and denominator in its factors and divide.

$66p{{q}^{2}}{{r}^{3}}=2\times 3\times 11\times p\times q\times q\times r\times r\times r$

$11q{{r}^{2}}=11\times q\times r\times r$

$66p{{q}^{2}}{{r}^{3}}\div 11q{{r}^{2}}=\dfrac{2\times 3\times 11\times p\times q\times q\times r\times r\times r}{11\times q\times r\times r}=6pqr$

iv. $34{{x}^{3}}{{y}^{3}}{{z}^{3}}\div 51x{{y}^{2}}{{z}^{3}}$

Ans: Write the numerator and denominator in its factors and divide.

$34{{x}^{3}}{{y}^{3}}{{z}^{3}}=2\times 17\times x\times x\times x\times y\times y\times y\times z\times z\times z$

$51x{{y}^{2}}{{z}^{3}}=3\times 17\times x\times y\times y\times z\times z\times z$

$34{{x}^{3}}{{y}^{3}}{{z}^{3}}\div 51x{{y}^{2}}{{z}^{3}}=\dfrac{2\times 17\times x\times x\times x\times y\times y\times y\times z\times z\times z}{3\times 17\times x\times y\times y\times z\times z\times z}$

$=\dfrac{2}{3}{{x}^{2}}y$

v. $12{{a}^{3}}{{b}^{8}}\div \left( -6{{a}^{6}}{{b}^{4}} \right)$

Ans: Write the numerator and denominator in its factors and divide.

$12{{a}^{8}}{{b}^{8}}=2\times 2\times 3\times {{a}^{8}}\times {{b}^{8}}$

$6{{a}^{6}}{{b}^{4}}=2\times 3\times {{a}^{6}}\times {{b}^{4}}$

$12{{a}^{8}}{{b}^{8}}\div \left( -6{{a}^{6}}{{b}^{4}} \right)=\dfrac{2\times 2\times 3\times {{a}^{8}}\times {{b}^{8}}}{-2\times 3\times {{a}^{6}}\times {{b}^{4}}}$

$=-2{{a}^{2}}{{b}^{4}}$

2. Divide the given polynomial by the given monomial.

i. $\left( 5{{x}^{2}}-6x \right)\div 3x$

Ans: Write the numerator its factors and divide.

$5{{x}^{2}}-6x=x(5x-6)$

$\left. 5{{x}^{2}}-6x \right)\div 3x=\dfrac{x(5x-6)}{3x}$

$=\dfrac{1}{3}(5x-6)$

ii. $\left( 3{{y}^{8}}-4{{y}^{6}}+5{{y}^{4}} \right)\div {{y}^{4}}$

Ans: Write the numerator its factors and divide.

$3{{y}^{8}}-4{{y}^{6}}+5{{y}^{4}}={{y}^{4}}\left( 3{{y}^{4}}-4{{y}^{2}}+5 \right)$

$\left( 3{{y}^{8}}-4{{y}^{6}}+5{{y}^{4}} \right)\div {{y}^{4}}=\dfrac{{{y}^{4}}\left( 3{{y}^{4}}-4{{y}^{2}}+5 \right)}{{{y}^{4}}}$

$=3{{y}^{4}}-4{{y}^{2}}+5$

iii.  $8\left( {{x}^{3}}{{y}^{2}}{{z}^{2}}+{{x}^{2}}{{y}^{3}}{{z}^{2}}+{{x}^{2}}{{y}^{2}}{{z}^{3}} \right)\div 4{{x}^{2}}{{y}^{2}}{{z}^{2}}$

Ans: Write the numerator its factors and divide.

$8\left( {{x}^{3}}{{y}^{2}}{{z}^{2}}+{{x}^{2}}{{y}^{3}}{{z}^{2}}+{{x}^{2}}{{y}^{2}}{{z}^{3}} \right)=8{{x}^{2}}{{y}^{2}}{{z}^{2}}(x+y+z)$

$8\left( {{x}^{3}}{{y}^{2}}{{z}^{2}}+{{x}^{2}}{{y}^{3}}{{z}^{2}}+{{x}^{2}}{{y}^{2}}{{z}^{3}} \right)\div 4{{x}^{2}}{{y}^{2}}{{z}^{2}}=\dfrac{8{{x}^{2}}{{y}^{2}}{{z}^{2}}(x+y+z)}{4{{x}^{2}}{{y}^{2}}{{z}^{2}}}$

$=2(x+y+z)$

iv. $\left( {{x}^{3}}+2{{x}^{2}}+3x \right)\div 2x$

Ans: Write the numerator its factors and divide.

${{x}^{3}}+2{{x}^{2}}+3x=x\left( {{x}^{2}}+2x+3 \right)$

$\left( {{x}^{3}}+2{{x}^{2}}+3x \right)\div 2x=\dfrac{x\left( {{x}^{2}}+2x+3 \right)}{2x}$

$=\dfrac{1}{2}\left( {{x}^{2}}+2x+3 \right)$

v. $\left( {{p}^{3}}{{q}^{6}}-{{p}^{6}}{{q}^{3}} \right)\div {{p}^{3}}{{q}^{3}}$

Ans: Write the numerator its factors and divide.

) ${{p}^{3}}{{q}^{6}}-{{p}^{6}}{{q}^{3}}={{p}^{3}}{{q}^{3}}\left( {{q}^{3}}-{{p}^{3}} \right)$

$\left( {{p}^{3}}{{q}^{6}}-{{p}^{6}}{{q}^{3}} \right)\div {{p}^{3}}{{q}^{3}}=\dfrac{{{p}^{3}}{{q}^{3}}\left( {{q}^{3}}-{{p}^{3}} \right)}{{{p}^{3}}{{q}^{3}}}$

$={{q}^{3}}-{{p}^{3}}$

3. Work out the following divisions.

i. $(10x-25)\div 5$

Ans: Write the numerator and denominator in its factors and divide.

$(10x-25)\div 5=\dfrac{2\times 5\times x-5\times 5}{5}$

$=\dfrac{5(2x-5)}{5}$

$=2x-5$

ii. $(10x-25)\div (2x-5)$

Ans: Write the numerator and denominator in its factors and divide.

$(10x-25)\div (2x-5)=\dfrac{2\times 5\times x-5\times 5}{(2x-5)}$

$=\dfrac{5(2x-5)}{2x-5}$

$=5$

iii. $10y(6y+21)\div 5(2y+7)$

Ans: Write the numerator and denominator in its factors and divide.

$10y(6y+21)\div 5(2y+7)=\dfrac{2\times 5\times y[2\times 3\times y+3\times 7]}{5(2y+7)}$

$=\dfrac{2\times 5\times y\times 3(2y+7)}{5(2y+7)}$

$=6y$

iv. $9{{x}^{2}}{{y}^{2}}(3z-24)\div 27xy(z-8)$

Ans: Write the numerator and denominator in its factors and divide.

$9{{x}^{2}}{{y}^{2}}(3z-24)\div 27xy(z-8)=\dfrac{9{{x}^{2}}{{y}^{2}}[3\times z-2\times 2\times 2\times 3]}{27xy(z-8)}$

$=\dfrac{xy\times 3(z-8)}{3(z-8)}$

$=xy$

v. $96abc(3a-12)(5b-30)\div 144(a-4)(b-6)$

Ans: Write the numerator and denominator in its factors and divide.

$96abc(3a-12)(5b-30)\div 144(a-4)(b-6)$

$=\dfrac{96abc(3\times a-3\times 4)(5\times b-2\times 3\times 5)}{144(a-4)(b-6)}$

$=\dfrac{2abc\times 3(a-4)\times 5(b-6)}{3(a-4)(b-6)}$

$=10abc$

4. Divide as directed.

i. $5(2x+1)(3x+5)\div (2x+1)$

Ans: Write the numerator and denominator in its factors and divide.

$5(2x+1)(3x+5)\div (2x+1)=\dfrac{5(2x+1)(3x+1)}{(2x+1)}$

$=5(3x+1)$

ii. $26xy(x+5)(y-4)\div 13x(y-4)$

Ans: Write the numerator and denominator in its factors and divide.

$26xy(x+5)(y-4)\div 13x(y-4)=\dfrac{2\times 13\times xy(x+5)(y-4)}{13x(y-4)}$

$=2$

iii. $52pqr(p+q)(q+r)(r+p)\div 104pq(q+r)(r+p)$

Ans: Write the numerator and denominator in its factors and divide.

$52pqr(p+q)(q+r)(r+p)\div 104pq(q+r)(r+p)$

$=\dfrac{2\times 2\times 13\times p\times q\times r\times (p+q)\times (q+r)\times (r+p)}{2\times 2\times 2\times 13\times p\times q\times (q+r)\times (r+p)}$

$=\dfrac{1}{2}r(p+q)$

iv. $20(y+4)\left( {{y}^{2}}+5y+3 \right)\div 5(y+4)$

Ans: Write the numerator and denominator in its factors and divide.

$20(y+4)\left( {{y}^{2}}+5y+3 \right)=2\times 2\times 5\times (y+4)\left( {{y}^{2}}+5y+3 \right)$

$20(y+4)\left( {{y}^{2}}+5y+3 \right)\div 5(y+4)=\dfrac{2\times 2\times 5\times (y+4)\times \left( {{y}^{2}}+5y+3 \right)}{5\times (y+4)}$                    $=4\left( {{y}^{2}}+5y+3 \right)$

v. $x(x+1)(x+2)(x+3)\div x(x+1)$

Ans: Write the numerator and denominator in its factors and divide.

$x(x+1)(x+2)(x+3)\div x(x+1)=\dfrac{x(x+1)(x+2)(x+3)}{x(x+1)}$

$=\left( x+2 \right)\left( x+3 \right)$

5. Factorise the expressions and divide them as directed.

i. $\left( {{y}^{2}}+7y+10 \right)\div (y+5)$

Ans: Factorise the given terms separately.

$\left( {{y}^{2}}+7y+10 \right)={{y}^{2}}+2y+5y+10$

$=y(y+2)+5(y+2)$

$=(y+2)(y+5)$

Divide the two terms.

$\left( {{y}^{2}}+7y+10 \right)\div (y+5)=\dfrac{(y+5)(y+2)}{(y+5)}$

$=y+2$

ii. $\left( {{m}^{2}}-14m-32 \right)\div (m+2)$

Ans: Factorise the given terms separately.

${{m}^{2}}-14m-32={{m}^{2}}+2m-16m-32$

$=m(m+2)-16(m+2)$

$=(m+2)(m-16)$

Divide the two terms.

$\left( {{m}^{2}}-14m-32 \right)\div (m+2)=\dfrac{(m+2)(m-16)}{(m+2)}$

$=m-16$

iii. $\left( 5{{p}^{2}}-25p+20 \right)\div (p-1)$

Ans: Factorise the given terms separately.

$5{{p}^{2}}-25p+20=5\left( {{p}^{2}}-5p+4 \right)$

=5[p(p-1)-4(p-1)]

$=5(p-1)(p-4)$

Divide the two terms.

$\left( 5{{p}^{2}}-25p+20 \right)\div (p-1)=\dfrac{5(p-1)(p-4)}{(p-1)}$

$=5(p-4)$

iv. $4yz\left( {{z}^{2}}+6z-16 \right)\div 2y(z+8)$

Ans: Factorise the given terms separately.

$4yz\left( {{z}^{2}}+6z-16 \right)=4yz\left[ {{z}^{2}}-2z+8z-16 \right]$

$=4\text{ }y\text{ }z\left[ z\left( z-2 \right)+8\left( z-2 \right) \right]$

$=4\text{ }y\text{ }z\left( z-2 \right)\left( z+8 \right)$

Divide the two terms.

$4yz\left( {{z}^{2}}+6z-16 \right)\div 2y(z+8)=\dfrac{4yz(z-2)(z+8)}{2y(z+8)}$

$=2z(z-2)$

v. $5pq\left( {{p}^{2}}-{{q}^{2}} \right)\div 2p(p+q)$

Ans: Factorise the given terms separately.

$5pq\left( {{p}^{2}}-{{q}^{2}} \right)=5pq(p-q)(p+q)$

Divide the two terms.

$5pq\left( {{p}^{2}}-{{q}^{2}} \right)\div 2p(p+q)=\dfrac{5pq(p-q)(p+q)}{2p(p+q)}$

$=\dfrac{5}{2}q(p-q)$

vi. $12xy\left( 9{{x}^{2}}-16{{y}^{2}} \right)\div 4xy(3x+4y)$

Ans: Factorise the given terms separately.

$12xy\left( 9{{x}^{2}}-16{{y}^{2}} \right)=12xy\left[ {{(3x)}^{2}}-{{(4y)}^{2}} \right]$

$=12xy(3x-4y)(3x+4y)$

Divide the two terms.

$12xy\left( 9{{x}^{2}}-16{{y}^{2}} \right)\div 4xy(3x+4y)$

$=\dfrac{2\times 2\times 3\times x\times y\times (3x-4y)\times (3x+4y)}{2\times 2\times x\times y\times (3x+4y)}$

$=3\left( 3\text{ }x-4\text{ }y \right)$

vii. $39{{y}^{3}}\left( 50{{y}^{2}}-98 \right)\div 26{{y}^{2}}(5y+7)$

Ans: Factorise the given terms separately.

$39{{y}^{3}}\left( 50{{y}^{2}}-98 \right)=3\times 13\times y\times y\times y\times 2\left[ \left( 25{{y}^{2}}-49 \right) \right]$

$=3\times 13\times 2\times y\times y\times y\times \left[ {{(5y)}^{2}}-{{(7)}^{2}} \right]$

$=3\times 13\times 2\times y\times y\times y(5y-7)(5y+7)$

$26{{y}^{2}}(5y+7)=2\times 13\times y\times y\times (5y+7)$

Divide the two terms.

$39{{y}^{3}}\left( 50{{y}^{2}}-98 \right)\div 26{{y}^{2}}(5y+7)=\dfrac{3\times 13\times 2\times y\times y\times y(5y-7)(5y+7)}{2\times 13\times y\times y\times (5y+7)}$

$=3y(5y-7)$

$=15{{y}^{2}}-21y$

## Overview of Deleted Syllabus for CBSE Factorisation Class 8 Maths Factorisation

 Chapter - 10 Dropped Topics Factorisation 12.4 - Division of Algebraic Expressions Continuation 12.5 Can you find the error?

## Class 8 Maths Chapter 12: Exercises Breakdown

 Chapter 12 -  Factorisation Exercises in PDF Format Exercise 12.1 3 Questions with Solutions (Short answer type) Exercise 12.2 5 Questions with Solutions (Short answer type) Exercise 12.3 5 Questions with Solutions (Short answer type)

## Conclusion

Chapter 12 of Class 8 Maths, "Factorisation," focuses on breaking down algebraic expressions into simpler factors, making it easier to solve equations. Key concepts include the factorisation of polynomials, using algebraic identities, and dividing algebraic expressions. It's important to understand how to identify common factors and apply formulas accurately. Students should focus on practicing different factorisation techniques and be aware of common errors such as incorrect application of identities and signs. Mastery of these basics is crucial for tackling more complex algebraic problems effectively by Vedantu.

## Other Study Material for CBSE Class 8 Maths Chapter 12

 S.No. Important Links for Chapter 12 Factorisation 1 Class 8 Factorisation Important Questions 2 Class 8 Factorisation Chapter Notes 3 Class 8 Factorisation Formula 4 Class 8 Factorisation RD Sharma Solutions 5 Class 8 Factorisation RS Aggarwal Solutions

## Chapter-Specific NCERT Solutions for Class 8 Maths

Given below are the chapter-wise NCERT Solutions for Class 8 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

## FAQs on NCERT Solutions for Class 8 Maths Chapter 12 Factorisation

1. Solve 49𝑦2 + 84𝑦𝑧 + 36𝑧2

49𝑦2 + 84𝑦𝑧 + 36𝑧2

49𝑦+ 84𝑦𝑧 + 36𝑧2

= (7𝑦)2+ 84𝑦𝑧 + (6𝑧)2

= (7𝑦)2 + 2 × 7𝑦 × 6𝑧 + (6𝑧)2

= (7𝑦)2+ (6𝑧)2+ 2 × 7𝑦 × 6𝑧

Using (𝑎 + 𝑏)2= 𝑎2 + 𝑏2 + 2𝑎𝑏

Here, 𝑎 = 7𝑦 and 𝑏 = 6𝑧

= (7𝑦 + 6𝑧)2

It is the solution.

2. Solve a4+ 2a2b2 + b4

Given expression is, 𝑎4 + 2𝑎2𝑏2 + 𝑏4

It is a fourth-degree expression.

= 𝑎4+ 2𝑎2𝑏2+ 𝑏4

Using the formula, (𝑎m)n = 𝑎m*n

∴ (𝑎2)2 = 𝑎2*2= 𝑎4

= (𝑎2)2+ 2𝑎2𝑏2+ (b2)2

= (𝑎2)2 + 2(𝑎2 × 𝑏2) + (𝑏2)2

= (𝑎2)2+ (𝑏2)2+ 2(𝑎2× 𝑏2)

Using (𝑥 + 𝑦)2  = x2 + y2 + 2xy, we can substitute as,

Here, 𝑥 = 𝑎2 and 𝑦 = 𝑏2

= (a2 + b2)2

Hence it is solved.

3. How to study Chapter 14 “Factorisation” of Class 8 Maths using Vedantu’s NCERT Solutions?

To use Vedantu's NCERT Solutions for Chapter 14 “Factorisation” of Class 8 Maths effectively for your studies:

• Read the NCERT textbook well and practice all of its exercise questions.

• Refer to Vedantu's NCERT Solutions of this chapter for perfectly crafted step-by-step answers.

• Use Vedantu's explanations provided below the solutions for an added understanding of the chapter.

• These explanations are also useful for revision purposes.

• You may also attend Vedantu's live classes to study the chapter with our master teachers.

4. What are the main topics covered in the NCERT Solutions for Chapter 14 of Class 8 Maths?

Chapter 14 "Factorisation" of Class 8 Maths contains the following foundational topics significant in various mathematical operations:

• Factors of natural numbers

• Factors of algebraic expressions

• Method of on factors

• Factorization by regrouping terms

• Factorization using identities

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

• Division of Algebraic Expressions

• Can you find the error?

5. How many exercises are there in NCERT Solutions for Chapter 14 of Class 8 Maths?

Chapter 14 “Factorisation” of Class 8 Maths contains a total of four exercises.

All four exercises are important from the examination point of view.

Factorization is a significant chapter and students must have the basics of this chapter clear in their minds. You can look up Vedantu's explanations of this chapter for further clarity.

You need to practice the exercises from this chapter diligently to master the conceptual understanding of this chapter. Refer to the NCERT Solutions for Chapter 14 of Class 8 Maths available free of cost on the Vedantu website and the Vedantu app.

6. What do you mean by factorization according to Chapter 14 of Class 8 Maths?

Factorization is the process of breaking down a number into smaller numbers which when multiplied together produces the original number.

In simple terms Factorisation of an algebraic expression is the process of writing the expression as a product of its factors. These factors may be numbers, algebraic variables, or algebraic expressions.

"Factorisation" is an important chapter as it lays down the basic foundations of several mathematical operations.

7. Explain the common factor method in factorization according to Chapter 14 of Class 8 Maths.

The Common Factor method is an organized way of factorizing an expression.

Here is how you can systematically factorize an expression using the common factor method:

• Write each term of the expression as a product of irreducible factors.

• Look for the common factors and separate them.

• Combine the remaining factors in each term in accordance with the distributive law.

Exercise 14.1 of Chapter 14 of Class 8 Maths is based on the common factor method. You can download the solutions for this exercise in pdf format here.

8. What is factorisation in class 8 chapter factorisation?

Factorisation is the process of breaking down a number or an algebraic expression into its simplest building blocks, called factors. These factors are numbers or expressions that, when multiplied together, give the original number or expression.Factorisation helps simplify complex problems and makes it easier to solve equations.

9. What are the different methods of factorisation in class 8th factorisation?

The different methods of factorisation in class 8th factorisation involve breaking down expressions into simpler parts. First, you can factor out the greatest common factor (GCF), which is the largest number or term that divides all the terms in the expression. Another method is grouping, where you group terms with common factors and then factor them. Special formulas, like the difference of squares and perfect square trinomials, can also be used for quick factorisation. Lastly, for quadratic expressions, you can split the middle term into two terms that add up to the original middle term and then factor by grouping.

10. How do you factorise a quadratic expression in Class 8 Maths Ch Factorisation?

To factorize a quadratic expression in class 8 chapter factorisation ,start by finding two numbers. These numbers need to do two things: multiply together to give the product of the first and last coefficients, and add up to the middle coefficient. Once you have these numbers, rewrite the middle term of the quadratic by splitting it according to these numbers. Then group the terms in pairs that share a common factor, pull out the common factor from each group, and you'll have your expression neatly factorized.

11. What are the key exercises in Class 8 Maths Ch Factorisation?

Class 8 Maths Ch Factorisation focuses on different factorization techniques, including factoring quadratics, using the greatest common factor, and special formulas like the difference of squares. These exercises help students break down complex expressions into simpler, manageable parts, crucial for solving various mathematical problems effectively. This practice is essential for mastering the foundational concepts of algebra.