Factorisation Class 8 Notes CBSE Maths Chapter 14 (Free PDF Download)

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;

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.