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NCERT Solutions for Class 8 Maths Chapter 12 Factorisation

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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.

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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 12.1 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 12.2 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


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.



Important Related Links for CBSE Class 8 Maths

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FAQs on NCERT Solutions for Class 8 Maths Chapter 12 Factorisation

1. What is factorisation in NCERT Class 8 Maths Chapter 12?

Factorisation is the process of expressing an algebraic expression as a product of its factors. These factors may be numbers, algebraic variables, or expressions that, when multiplied together, re-create the original expression. In Class 8 Maths Chapter 12, factorisation is essential for simplifying and solving algebraic problems as per the CBSE 2025–26 syllabus.

2. Which methods of factorisation are taught in Class 8 Maths Chapter 12 as per NCERT Solutions?

The main methods of factorisation covered in Class 8 Chapter 12 are:

  • Taking out the greatest common factor (GCF)
  • Factorisation by grouping terms
  • Using algebraic identities (such as (a+b)2 and a2-b2)
  • Factorisation of quadratic expressions by splitting the middle term
Students are encouraged to practice and apply all methods for various types of questions.

3. How do you apply the common factor method in factorisation according to NCERT Class 8 guidelines?

To use the common factor method:

  • Write each term as a product of its prime (or irreducible) factors
  • Identify the common factors among all terms
  • Take out the common factor and write the expression as a product
This method is especially important for the initial exercises of Chapter 12 and helps simplify polynomials efficiently.

4. How can students avoid common mistakes when using algebraic identities in factorisation?

To avoid mistakes with algebraic identities:

  • Memorize standard identities thoroughly (e.g., (a+b)2 = a2 + 2ab + b2)
  • Check signs and coefficients carefully in each step
  • Confirm the expanded form matches the original expression
  • Always write down intermediate steps to catch calculation errors
Accurate application of identities is crucial for scoring marks in CBSE 2025–26 exams.

5. What steps should be followed to factorise a quadratic expression in Class 8 Maths Ch 12?

For factorising a quadratic expression of form ax2+bx+c:

  • Find two numbers that multiply to a×c and add to b
  • Rewrite the middle term b as the sum of these two numbers
  • Group the terms in pairs and factor out the common factor from each
  • Express as a product of two binomials
This method is central to Exercise 12.2 and 12.3 in the latest NCERT textbook.

6. Why is mastering factorisation crucial for success in higher algebra topics?

Mastering factorisation builds the foundation for advanced algebra topics such as equations, polynomials, and algebraic fractions. It enables students to simplify expressions, solve equations efficiently, and understand the structure of more complex problems encountered in higher classes and competitive exams.

7. What are the main types of expressions that students learn to factorise in NCERT Class 8 Maths Chapter 12?

Students learn to factorise:

  • Monomials and binomials
  • Quadratic trinomials
  • Expressions using identities (a2-b2, perfect squares, etc.)
  • Difficult algebraic expressions through grouping or splitting
Practice on all these types aids exam readiness as per the revised 2025–26 syllabus.

8. How is the division of algebraic expressions introduced and tested in Class 8 factorisation exercises?

Division is covered by teaching how to divide one algebraic expression by another (especially monomial by monomial, polynomial by monomial). Students must:

  • Express numerator and denominator in factorised form
  • Simplify by cancelling common factors where possible
  • Obtain the final reduced expression
These problems feature in Exercise 12.3 of the chapter.

9. What is the role of factorisation in solving equations in Class 8 Maths?

Factorisation allows students to break down complex expressions into simpler products, which makes solving equations easier. By factorising expressions and equating them to zero, students can find the roots (solutions) systematically. This skill connects directly with the linear equations chapter and is fundamental for CBSE examination questions.

10. What topics or questions have been deleted from the Class 8 Maths Chapter 12 syllabus for 2025–26?

According to the official CBSE curriculum update, topics 12.4 (division of algebraic expressions continuation) and 12.5 (Can you find the error?) have been dropped from Class 8 Maths Chapter 12 for the 2025–26 session.

11. How can students practice and master factorisation as per CBSE recommendations for Class 8?

Students are advised to:

  • Go through all solved examples in the NCERT textbook
  • Solve every exercise problem without skipping steps
  • Review mistakes and revisit tricky identity-based questions
  • Use NCERT Solutions for stepwise guidance, as per CBSE pattern
  • Attend classes and clarify doubts with teachers
Regular practice ensures conceptual clarity and exam confidence.

12. What common misconceptions might students have about factorisation in Class 8, and how can they avoid them?

Common misconceptions include:

  • Confusing factorisation with expansion (opposite process)
  • Missing out non-numeric common factors
  • Neglecting negative signs or coefficients
  • Incorrect use of identities for non-applicable expressions
To avoid these, students must follow each factorisation step carefully and double-check work, especially with identities or multiple terms.

13. What is the importance of solving all exercises in NCERT Solutions for Class 8 Maths Chapter 12?

Solving all exercises ensures full exposure to all types of factorisation problems covered in the CBSE syllabus. It prepares students for variations in exam questions, helps in error spotting, and builds speed and accuracy required for board tests and school exams.

14. How does factorisation help in simplifying algebraic fractions?

Factorisation breaks complex algebraic fractions into simpler units by cancelling common factors in numerator and denominator. This process makes further calculations and comparisons much easier in algebraic operations, as taught in Class 8 Chapter 12.

15. Why should students pay attention to CBSE-updated exercises and solutions for Chapter 12 Factorisation?

CBSE updates ensure that students are studying only the topics relevant for current academic years. Referring to the latest exercises and solutions helps avoid unnecessary study of deleted portions and keeps focus on scoring areas strongly aligned with board exam requirements.