# Factorisation

When we factorise an algebraic expression, we write it as a product of factors. These factors may be numbers, algebraic variables or algebraic expressions.

Expressions like 4xy, 6x²y , 3x (y + 2), 7(y + 1) (x + 5) are already in factor form.

Consider expressions like 2x + 4, 3x + 3y, x2 + 5x, x2 + 5x + 6.

It is not obvious what their factors are. We need to develop systematic methods to factorise these expressions, i.e., to find their factors.

## Method of common factors

Example 1:

Factorise 2x + 4.

Step 1: Write each term as a product of irreducible factors;

2x = 2 × x

4 = 2 × 2

Hence 2x + 4 = (2 × x) + (2 × 2)

Step 2: Note that factor 2 is common to both the terms.

Step 3: By distributive law

2 × (x + 2) = (2 × x) + (2 × 2)

Step 4: Therefore,

2x + 4 = 2 × (x + 2) = 2 (x + 2)

Thus, the expression 2x + 4 is the same as 2 (x + 2). Now we can read off its factors:

they are 2 and (x + 2). These factors are irreducible.

Example 2:

Factorise 6xy + 12x.

The irreducible factor forms of 5xy and 10x are respectively,

6xy = 6 × x × y

12x = 2 × 6 × x

Observe that the two terms have 6 and x as common factors. Now,

6xy + 12x = (6 × x × y) + (6 × x × 2)

= (6x × y) + (6x × 2)

We combine the two terms using the distributive law,

(6× y) + (6x× 2) = 6x × ( y + 2)

Therefore, 6xy + 12x = 6 x (y + 2).

## Factorisation by regrouping terms

Let us consider the following expression

2xy + 2y + 3x + 3.

Observe that first two terms have common factors 2 and y and the last two terms have a common factor 3. But there is no single factor common to all the terms.

Now we need to think how to proceed.

Let us write (2xy + 2y) in the factor form:

2xy + 2y = (2 × x × y) + (2 × y)

= (2 × y × x) + (2 × y × 1)

= (2y × x) + (2y × 1) = 2y (x + 1)

Similarly, 3x + 3 = (3 × x) + (3 × 1)

= 3 × (x + 1) = 3 ( x + 1)

Hence, 2xy + 2y + 3x + 3 = 2y (x + 1) + 3 (x +1)

Observe, now we have a common factor (x + 1) in both the terms on the right hand

side. Combining the two terms,

2xy + 2y + 3x + 3 = 2y (x + 1) + 3 (x + 1) = (x + 1) (2y + 3)

The expression 2xy + 2y + 3x + 3 is now in the form of a product of factors. Its

factors are (x + 1) and (2y + 3).

We also understand that these factors are irreducible.

## Regrouping

Rearranging the expression in order to factorise it is called regrouping. For easy understanding, let us consider the above expression  2xy + 3 + 2y + 3x;

Rearranging the expression, as 2xy + 2y + 3x + 3, allows us to form groups (2xy + 2y) and (3x + 3) leading to factorisation. This is regrouping.

Regrouping may be possible in more than one ways. Suppose, we regroup the

expression as: 2xy + 3x + 2y + 3. This will also lead to factors. Let us try:

2xy + 3x + 2y + 3 = 2 × x × y + 3 × x + 2 × y + 3

= x × (2y + 3) + 1 × (2y + 3)

= (2y + 3) (x + 1)

The factors are the same (as they have to be), although they appear in different order.

Example 3: Factorise 6xy – 4y + 6 – 9x.

## Solution:

If there is no common factor among all terms, we can proceed by regrouping the expression.

We observe that first two terms have a common factor 2y;

6xy – 4y = 2y (3x – 2) (a)

If you change the order of last two terms as – 9x + 6, we get the factor ( 3x – 2)

–9x + 6 = –3 (3x) + 3 (2)

= – 3 (3x – 2) (b)

By Putting expressions (a) and (b) together,

6xy – 4y + 6 – 9x = 6xy – 4y – 9x + 6

= 2y (3x – 2) – 3 (3x – 2)

= (3x – 2) (2y – 3)

The factors of (6xy – 4y + 6 – 9 x) are (3x – 2) and (2y – 3).

## Factorisation using identities

We know that (a + b)2 = a2 + 2ab + b2 (I)

(a – b)2 = a2 – 2ab + b2 (II)

(a + b) (a – b) = a2 – b2 (III)

Let us try to solve an example using the identities given above.

Example : Factorise x2 + 8x + 16

## Solution

The expression; it has three terms. Therefore, it does not fit Identity III.

First and third terms are perfect squares with a positive sign before the middle term. So, it is of the form a² + 2ab + b² where a = x and b = 4

such that a2 + 2ab + b2 = x2 + 2 (x) (4) + 42

= x2 + 8x + 16

Since a2 + 2ab +b2

## Solution

The expression; it has three terms. Therefore, it does not fit Identity III.

First and third terms are perfect squares with a positive sign before the middle term. So, it is of the form a² + 2ab + b² where a = x and b = 4

such that a2 + 2ab + b2 = x2 + 2 (x) (4) + 42

= x2 + 8x + 16

Since a2 + 2ab +b2