Factorisation by grouping the terms

Take the expression:

2xy + 2y + 3x + 3 as an example

What do we observe? Is there something different about this expression from the ones that we have considered upto this point?

We observe that the terms all together don't share a common factor. The first two terms have the common factors of 2 and y while the last two terms have a common factor of 3. So, how to solve this expression now?

Let's break the expression into smaller pieces and let's just consider the terms '2xy' and '2y'

Writing the term (2xy + 2y) in the factor form, we get:

2xy + 2y =

Similarly we can take the terms (3x + 3) which will give us:

3x + 3 =

Hence combining the results we can write:

2xy + 2y + 3x + 3 = (Take out the common factors and put remaining in brackets)

We notice that the term (x + 1) is common to both the terms on the right hand side of the equation. Upon bringing out the common term we will get:

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

2xy + 2y + 3x + 3 =

The expression 2xy + 2y + 3x + 3 is now in the form of the product of its factors with the factors being (x + 1) and (2y + 3). Note that these factors are also irreducible.

What is regrouping?

As we just saw in the above expression:

2xy + 3 + 2y + 3x

it is not immediately evident what the possible factors may be. In order to further simplify this process we will rearrange the expression as

2xy + 2y + 3x + 3

thus, allowing us to form the groups (2xy + 2y) and (3x + 3) leading to factorisation. This is known as regrouping.

Regrouping may be possible in more than one ways. Suppose, we regroup and rewrite the expression as:

2xy + 3x + 2y + 3

This will also lead to us getting the factors:

2xy + 3x + 2y + 3 = x × ( ) + 1 × ( )

This gives us

=

Notice that the factors are the same (as they should be) even though they appear in different order.

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

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