# Methods of Factorisation

1. Method of Common Factors

• In this method, we have to write the irreducible factors of all the terms
• Then find the common factors amongst all the irreducible factors.
• The required factor form is the product of the common term we had chosen and the leftover terms.

2. Factorisation by Regrouping Terms

When there is no common term in the expressions then

• We have to make the groups of the terms.
• Then choose the common factor among these groups.
• Find the common binomial factor and it will give the required factors.

3. Factorisation Using Identities

Remember some identities to factorise the expression

• (a + b)2  = a2 + 2ab + b2
• (a – b)2  = a2 – 2ab + b2
• (a + b) (a – b)  = a2 – b2

(2x + 3)2 = (2x)2+ 2(2x) (3) + (3)2 = 4x2 + 12x + 9

(2x – 3)2 = (2x)2 – 2(2x)(3) + (3)2 = 4x-12x + 9

(2x + 3) (2x – 3) = (2x)2 – (3)2 = 4x2 – 9

Example 1

Factorise x2– (2x – 1)2 using identity.

Solution:

This is using the identity (a + b) (a – b) = a2 – b2

x2– (2x – 1)2 = [(x + (2x – 1))] [x – (2x-1))]

= (x + 2x – 1) (x – 2x + 1)

= (3x – 1) (- x + 1)

Example 2

Factorize 9x² – 24xy + 16y² using identity.

Solution:

9x² – 24xy + 16y²

= (3x)2 – 24xy + (4y)2

Now split the middle term.

= (3x)2 – 2(3x) (4y) + (4y)2

=(3x – 4y)2

Hence the factors are (3x – 4y) (3x – 4y).

Example 3

Factorise x2 + 10x + 25 using identity.

Solution:

x2 + 10x + 25

= (x)2 + 2(5) (x) + (5)2

x2 + 10x + 25 = (x + 5)2

4. Factors of  the form ( x + a) ( x + b)

(x + a) (x + b) = x2 + (a + b) x + ab.

Example:

Factorise x2 + 3x + 2.

Solution:

If we compare it with the identity (x + a) (x + b) = x2 + (a + b) x +ab

We get to know that (a + b) = 3 and ab = 2.

This is possible when a = 1 and b = 2.

Substitute these values into the identity,  Scroll to Top