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# Factorisation Class 8 Notes CBSE Maths Chapter 12 (Free PDF Download)

Last updated date: 07th Sep 2024
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## Revision Notes for CBSE Class 8 Maths Chapter 12 - Free PDF Download

Free PDF download of Class 8 Maths Chapter 12 - Factorisation Revision Notes & Short Key-notes prepared by expert Maths teachers from latest edition of CBSE(NCERT) books. To register Maths Tuitions on Vedantu.com to clear your doubts. You can also register Online for Class 8 Science tuition on Vedantu.com to score more marks in CBSE board examination. Vedantu is a platform that provides free CBSE Solutions (NCERT) and other study materials for students. Maths Students who are looking for the better solutions ,they can download Class 8 Maths NCERT Solutions to help you to revise complete syllabus and score more marks in your examinations.

Also, check CBSE Class 8 Maths revision notes for All chapters:

## Access Class 8 Mathematics Chapter 12 – Factorisation Notes in 30 Minutes

Factorisation:

Factorisation of a number or an algebraic expression is writing it as a product of numbers, variables or expressions. These numbers, variables, or expressions are called factors.

• Factorisation of numbers – A number can be broken down as a product of two or more factors depending upon the number given. For example, the number $15$ can be written as the product of $1$ and $15$ or the product of $3$ and $5$ or the product of $1,3$ and $5$. So, $1,15,3$ and $5$ are the factors of the number $30$. Here, if the factors are prime numbers, then they are called prime factors.

• Factorisation of algebraic expressions – The factors can be a number, variable or an expression. For example, factors of $5ab$ are$5,a$ and $b$. Similarly, the expression $3ab+6a$ can be written as $3ab+6a=3\times a\times (b+2)$, where $3,a$ and $(b+2)$ are the factors.

Method Of Common Factors:

In this method of finding the factors of an algebraic expression, we first write each term of an expression as a product of irreducible factors. Then by using distributive law, we take the common factors and thus determine the factors of that algebraic expression. For example, let us find the factors of the expression $4x+2xy$. First, find the factors of each term.

So, $4x$ can be written as $4x=2\times 2\times x$ and $2xy$ can be written as $2\times x\times y$. Thus, we can write $4x+2xy=(2\times 2\times x)+(2\times x\times y)$, where it can be noticed that $2$ and $x$ are common in both terms. So, $4x+2xy=2\times x\times (2+y)$. Thus, the common factors are $2,x$ and $(2+y)$.

Factorisation By Regrouping Terms:

When the terms of a given expression do not have any single common factor, then we rearrange the terms to group them to lead to factorisation. This method of forming groups is called regrouping. There can be one or more ways of regrouping for a given expression. For example, take the expression $z-7+7xy-xyz$, here the terms do not have any common factors, so we will rearrange the terms as $z-xyz-7+7xy$ and then taking common $z$ and $-7$ from first the two and last two terms respectively. We get;

$z-xyz-7+7xy=z\times \left( 1-xy \right)-7\times \left( 1-xy \right)$

$=\left( z-7 \right)\times (1-xy)$

Hence, the factors are $\left( z-7 \right)$ and $(1-xy)$.

Factorisation Using Identities:

The expressions directly or indirectly given in certain forms can be factored using standard identities. Some common identities and their expressions have been tabulated below;

 Expressions Of The Form Identity For Factorisation ${{a}^{2}}+2ab+{{b}^{2}}$ ${{\left( a+b \right)}^{2}}$ ${{a}^{2}}-2ab+{{b}^{2}}$ ${{\left( a-b \right)}^{2}}$ ${{a}^{2}}-{{b}^{2}}$ $(a+b)(a-b)$ ${{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc+2ac$ ${{\left( a+b+c \right)}^{2}}$ ${{a}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}+{{b}^{3}}$ ${{\left( a+b \right)}^{3}}$ ${{a}^{2}}+(x+y)a+xy$ $\left( a+x \right)(a+y)$

Apart from the above identities, many other identities are also used. The signs of the terms should be specially taken care of while using the identities.

Division Of Algebraic Expressions:

The division operation in the case of algebraic expressions unlike numbers is a bit different and factorisation plays an important role in it. The division is carried out keeping factorisation as a key player, which will see in the following cases. Note that the examples considered upon division give $0$ as their remainder.

• Division of a monomial by another monomial – Since monomials have only one term, it is written as products of its irreducible factors and then the common factors in two monomials are then canceled. Let us suppose we need to divide $14{{x}^{2}}y{{z}^{3}}$ by $7xyz$. We can write this as;

$\dfrac{14{{x}^{2}}y{{z}^{3}}}{7xyz}=\dfrac{2\times 7\times x\times x\times y\times z\times z\times z}{7\times x\times y\times z}$

$=2x{{z}^{2}}$

Hence, the quotient is $2x{{z}^{2}}$.

• Division of a polynomial by a monomial – In this case, each term of the polynomial is separately divided by the monomial. Hence, each term is written into its factor form and then divided separately. For example, let us divide the polynomial $4{{a}^{3}}+2{{a}^{2}}+6a$ by monomial $2a$. So, first we can write the polynomial as;

$4{{a}^{3}}+2{{a}^{2}}+6a=(2\times 2\times a\times a\times a)+\left( 2\times a\times a \right)+\left( 2\times 3\times a \right)$

and the monomial as $\left( 2\times a \right)$.

Hence, $\dfrac{4{{a}^{3}}+2{{a}^{2}}+6a}{2a}=\dfrac{(2\times 2\times a\times a\times a)+\left( 2\times a\times a \right)+\left( 3\times 2\times a \right)}{\left( 2\times a \right)}$

$=\dfrac{\left( 2\times 2\times a\times a\times a \right)}{\left( 2\times a \right)}+\dfrac{\left( 2\times a\times a \right)}{\left( 2\times a \right)}+\dfrac{\left( 3\times 2\times a \right)}{\left( 2\times a \right)}$

$=\left( 2\times a\times a \right)+\left( a \right)+\left( 3 \right)$

$=2{{a}^{2}}+a+3$

• Division of a polynomial by another polynomial – In this case also, both numerator and denominator are factored and the factors common in both are cancelled. For example, let us divide $\left( 7{{x}^{2}}+14x \right)$ by $\left( x+2 \right)$, then we can write;

$\dfrac{\left( 7{{x}^{2}}+14x \right)}{\left( x+2 \right)}=\dfrac{\left( 7\times x\times x \right)+\left( 2\times 7\times x \right)}{\left( x+2 \right)}$

Take $7x$ common from the two terms of numerator,

$=\dfrac{7\times x\times \left( x+2 \right)}{\left( x+2 \right)}=7x$

Hence, the quotient is $7x$.

Common Errors While Solving Exercises:

• There might be terms in an algebraic expression without any coefficient, such terms have $1$ as their coefficient. So, we need to take care about that while doing addition or subtraction.

• While substituting negative values in an expression, always use brackets so as to avoid confusion in final signs.

• When we multiply an expression enclosed in a bracket by a constant or a variable or an expression, we should multiply it with each term of the expression that is within brackets.

• Whenever we raise the power of a monomial, the power of both constant and variable is needed to be raised.

## What are the Benefits of Referring to Vedantu’s Revision Notes for Class 8 Maths Chapter 12 - Factorisation

• Provides quick, clear summaries of key concepts.

• Simplifies complex topics for better understanding.

• Efficient tool for last-minute exam prep.

• Enhances retention of crucial information.

• Supports effective exam preparation with key points and tips.

• Saves time by consolidating information.

• Prioritizes important topics and questions.

• Offers practical examples for real-world connections.

• Boosts student confidence for exams.

## Conclusion

For an enhanced comprehension of this subject, NCERT - Class 8 Maths Chapter 12 - Factorisation thoughtfully prepared by experienced educators at Vedantu is your invaluable companion. These notes break down the complexities of Factorisation into easily digestible sections, helping you grasp new concepts, master formulas, and navigate through questions effortlessly quickly in the last minute as well. By immersing yourself in these notes, you not only prepare for your studies more efficiently but also develop a profound understanding of the subject matter.

## FAQs on Factorisation Class 8 Notes CBSE Maths Chapter 12 (Free PDF Download)

1. Are NCERT Solutions for Class 8 Maths Chapter 12 important from the exam point of view?

Yes, NCERT solutions of class 8 Maths Chapter 12 is important from an exam point of view as these questions will help you in saving your time and you will get accurate answers to all the questions. You can prepare from those questions a day before your exam and cross-check all the solutions so that you do not write wrong answers in your exams. The solutions are accurate and you can rely on them. You can find these solutions easily on Vedantu.

2. What are the main topics covered in the NCERT Solutions for Class 8 Maths Chapter 12?

The main topics covered in NCERT Solution are :

• Introduction of the chapter

• Factorization’s definition.

• Algebraic expression’s division

• Can you find the error

For more information, visit Vedantu's official website. There you will find the solved answers which will be helpful in your exams. These solutions will cover all the important topics which are curated together from the exam's point of view. You can download these questions and save them on your PC for future reference.

3. Why is learning Factorisation important from ncert solutions?

Class 8 Maths Chapter 12 NCERT Solutions Factorization encompasses all aspects of factorization and its applications, including subtopics. Algebraic expression factorization is a fundamental ability for advanced math courses. Factorization of expressions is a difficult operation that necessitates a thorough understanding of factors and how to list them step by step.

Find the common factors of the given terms.

(i) 12x, 36

The Factors of 12x and 36

12x = 2×2×3×x

36 = 2×2×3×3

We get the common factors of 12x and 36 that are 2, 2, 3

and , 2×2×3 = 12

(ii) 2y, 22xy

The Factors of 2y and 22xy

2y = 2×y

22xy = 2×11×x×y

We get the common factors of 2y and 22xy are 2, y

and,2×y = 2y

4. How do you learn Factorisation?

You can learn factorization by solving the questions from the NCERT book. Practice is the key to master factorization. From Vedantu, you can download the question paper and can practice the questions to know the pattern. There are solved questions available on the website. You can download that too and save it on your PC to use in the future. The questions which you will practice from the previous year's question paper will be helpful for you to solve questions that are based on similar patterns.

5. Factorise: x2 – 64.

x2 – 64

⇒ x2 – 82 [Since 8 x 8 = 64]

According to the formula,

⇒ a2 – b2 = (a + b) (a – b)

⇒ x2 – 82 = (x + 8) (x – 8)

For more information, visit Vedantu's official website(vedantu.com). There you will find the solved answers which will be helpful in your exams, for free. These solutions will cover all the important topics which are put together from the exam's point of view. The help could also be acquired from the Vedantu Mobile app.