Factorisation with Examples and FAQs

 

What is factorisation?

Factorisation is a method that helps to find the factors of numbers or mathematical expressions. It is defined as dividing or breaking an entity into a product of several smaller entities which are of the same type, i.e., numbers, algebraic terms, etc.

Factorisation of Numbers

Any number is a factor of a number if it divides the number without leaving any remainder behind. For example, 2 is a factor of 4, 5 is a factor of 20 hence, the factorisation of 20 = 4 × 5.

Factorisation of Algebraic Expressions

Now that we know what factorisation is, so it is easier to understand the factorisation of algebraic expressions. In this case, the expression is reduced to a simpler form, and are represented as a product of its factors. The factors of an algebraic expression are either integers, variables, or a simpler algebraic expression.

For example, the factorisation of the expression 3 x^{2}+9 x-30 \text { is }

3 x^{2}+9 x-30=3(x+5)(x-2). Here, 3, (x+5)\text{ and }(x-2)  are the factors of the expression.

Types of Factorisations

There are three ways using which an expression can be factorised, these are:

• Common Factor Method
• Regrouping terms Method
• Factoring using Identities

Let’s discuss each method in detail.

Common Factor Method

Let us consider an algebraic expression:

y^{3}+2 y^{2}

Here y^{3} \text { can be factorised as } y \times y \times y.

2 y^{2} \text { can be factorised as } 2 \times y \times y \text {. }

The Greatest common factor of the two terms is y^{2}.

Hence, the factorised expression is y^{2}(y+2).

This is the common factor method.

Regrouping Factor Method

For some cases all the terms in the expression may not have a single common factor, here the regrouping method comes to use. Let us consider an algebraic expression ( 28x + z + 2xz + 14 ).

Identify which terms have common factors. Here the first and the last term has a common factor 14. The second and third terms have a common factor z.

Thus, we regroup the terms:

28x + z + 2xz + 14 = ( 28x + 14 ) + ( z + 2xz ) 

Now following the method of common factor:

( 28x + 14 ) + ( z + 2xz ) = 14( 2x + 1 ) + z( 2x + 1 ).

Clearly ( 2x + 1 ) is a common factor, hence the factorised expression is:

( 2x + 1 )( 14 + z ) which can be written as ( 2x + 1 )( z + 14 ).

Factoring using Identities

List of common Identities

The list of common identities that are used for factorisation is:

1. (a+b)^{2}=a^{2}+2 a b+b^{2}

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

3. a^{2}-b^{2}=(a+b)(a-b)

4. (a+b)^{3}=a^{3}+3 a^{2} b+3 a b^{2}+b^{3}

5. (a-b)^{3}=a^{3}-3 a^{2} b+3 a b^{2}-b^{3}

6. a^{3}+b^{3}=(a+b)\left(a^{2}-a b+b^{2}\right)

7. a^{3}-b^{3}=(a-b)\left(a^{2}+a b+b^{2}\right)

For example, 4 y^{2}+4 y+1 is an algebraic expression, however, there are no common factors for the three terms in the expression. In this case, we check if the expression is similar to an identity to factorize easily.

4 y^{2}+4 y+1=(2 y)^{2}+2 \cdot(2 y) \cdot 1+1^{2}

Which is similar to the expression  (a+b)^{2}=a^{2}+2 a b+b^{2}.

Comparing the two expressions gives us the factors of the expression 4 y^{2}+4 y+1.

4 y^{2}+4 y+1=(2 y+1)^{2}=(2 y+1)(2 y+1)

\therefore \text { the factor is }(2 y+1)^{2} \text { or the factors are }(2 y+1) \text { and }(2 y+1) \text {. }

Examples

Example 1: Determine the factors of the algebraic expression a^{4}+6 a^{3}+9 a^{2}

Solution: 

The given expression is a^{4}+6 a^{3}+9 a^{2}.

The common factor in all the three terms is a^2, hence the expression reduces to:

a^{4}+6 a^{3}+9 a^{2}=a^{2}\left(a^{2}+6 a+9\right)

The part in the parenthesis can be factorised further using identity:

(x+b)^{2}=x^{2}+2 x b+b^{2}

a^{2}+6 a+9=a^{2}+2 \cdot(a) \cdot(3)+(3)^{2}=(a+3)^{2}

a^{4}+6 a^{3}+9 a^{2}=a^{2}(a+3)^{2}

The factors of the expression are a, a, (a+3),(a+3).

Example 2: Are 2, ( z + 2 ), ( z – 3 ) the factors of 2 z^{2}-2 z-12 ?

Solution:

The given terms are 2,(z+2) \text { and }(z-3). To check if they are the factors or not we must calculate the product of the terms.

2 \times(z+2) \times(z-3)=2[(z+2)(z-3)]=2\left[z^{2}-3 z+2 z-6\right]

\text { Now, } 2\left[z^{2}-3 z+2 z-6\right]=2\left[z^{2}-z-6\right]=2 z^{2}-2 z-12

\text { Hence, } 2,(z+2) \text { and }(z-3) \text { are the factors of } 2 z^{2}-2 z-12 \text {. }