This section exploles factorising linear expressions, it covers factorising by a common number: 4x+8y = 4(x+4y), factorising by a common letter, 4x+x² = x(4+x), factorising by a common number and letter 2x+4x² = 2x(1+2x).

Factorising means reverse of brackets. We put an expression back into brackets.

## Basic Factorisation

Example: Factorise the following expression;

6x + 24

Answer:

6( )

6(x + 4)

Explanation: Both 6 and 24 appear in the 6 times table, which means we simplify both and multiply by 6. This can be done with the help of brackets as shown below.
That means;

6( )

Since brackets means multiplication we will be multiplying with whatever is inside the brackets by 6. To find out what numbers will be in the brackets we must divide each term in the original expression by 6 because as you already know the reverse of multiplication is division.
6x ÷ 6 = x

That means;
6(x + )

24 ÷ 6 = 4

6(x + 4)

We have managed to factorise the original expression. You can test whether the answer is right by expanding the brackets.
Example: Factorise the following expression;

6a + 9b – 15c

Answer:

3( )

3(2a + 3b – 5c)

Explanation:
The first step we take is recognising that each term in the expression above can be divided by a number. This number is a factor of 6, 9, and -15, that’s why we say factorise. The highest common factor of these numbers is a 3.
Step 1 would be to setup the brackets, since 3 is the highest common factor we put this outside as shown below.
This gives the factorised expression as;

3( )

Step 3 we divide each term in the original expression;
6a ÷ 3 = 2a

9b ÷ 3 = 3b

-15c ÷ 3 = -15c

3(2a + 3b – 5c)

## Harder expressions

In the examples above we have only taken out a single term as a factor, below is an example where more numbers and letters are taken out.
Example: Factorise 4y – 6y² + 2y

^{3}
Answer:

2y( )

2y(2 – 3y + y²

2y(2 – 3y)

Explanation: First we notice that each term in the expression can be divided by
That means;
That means;

**y**and**2**. That means 2y is the factor for all the terms, so we take that outside the brackets as follows;2y( )

Step 2 we take each term in the expression in turn as divide by 2y.
2y × 2 = 4y

That means;
2y(2 )

2y × -3y = 6y²

2y(2 – 3y)

2y × y² = 2y³

2y(2 – 3y + y²

The above is the final factorised expression.