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Example

Factorise 6t + 10.

To factorise, look for a number which is a factor of both 6 and 10 (that is why it is called ‘factorising’).

Two is a factor of both numbers so 2 goes in front of the bracket.

��(2(…�Ħ�Ħ\)

Divide 6t by 2 to get the first term inside the bracket.

\(2(3t ……\)

The + does not change and the last term is \(10 ÷ 2 = 5\)

\(2(3t + 5)\)

Check that this is correct by multiplying out.

\(2(3t + 5) = 6t + 10✓\)

To factorise fully, it is important to find the highest common factor (HCF) of the terms.

Example

Factorise this expression:

\(54 – 18x + 36y\)

First find the HCF of 54, 18 and 36 which is 18.

This means that 18 goes in front of the bracket.

The terms inside the bracket are found by dividing each one by 18.

\(54 – 18x + 36y = 18(3 – x + 2y)\)

Check that this is correct by multiplying out.

\(18(3 – x + 2y) = 54 – 18x + 36y ✓\)

If we had used a smaller factor of 54, for example 9, the expression would not be fully factorised.

When factorising, always use the HCF of the terms.

Example

Factorise this expression:

\(x^2 – 2x\)

First find the HCF of \(x^2\) and \(2x\) which is \(x\).

This means that \(x\) goes in front of the bracket.

The terms inside the bracket are found by dividing each one by \(x\).

\(x^2 – 2x = x(x – 2)\)

Check that this is correct by multiplying out.

\(x(x – 2) = x^2 – 2x ✓\)

Example

Factorise the expression \(6mn + 4m^2n – 2mn^2\)

In the expression, there are 3 different types of terms – numbers, terms in m and terms in n.

We need to find the HCF for each of these.

Numbers: HCF of 6, 4 and 2 is 2.

Terms in m: HCF is m.

Terms in n : HCF is n.

HCF for the expression is 2mn.

\(6mn + 4m^2n – 2mn^2 = 2mn (3 + 2m – n)\)

Check that this is correct by multiplying out.

\(2mn (3 + 2m – n) = 6mn + 4m^2n – 2mn^2 ✓\)