One common factor

To factorise an expression, we need to take out any factors that are common to each term. The process is the opposite of expanding brackets.

To make sure an expression is fully factorised, we need to identify its (HCF), ie the biggest number/letter that each term can be divided by. Otherwise the expression will not be fully factorised, because there will still be common factors inside the bracket.

Example one

Factorise \(\text{10x + 25}\)

We need to find the HCF of \(\text{10x}\) and \(\text{25}\). The biggest number both terms can be divided by is 5, so the HCF is 5.

This is what goes outside the bracket: \(\text{5(? + ?)}\)

To identify the terms that need to go inside the bracket, we must divide each term by the highest common factor.

\(\text{10x ÷ 5 = 2x}\)

\(\text{25 ÷ 5 = 5}\)

So we have \(\text{5(2x + 5)}\)

We can check if our answer is correct by expanding the bracket.

\(\text{5 × 2x = 10x}\)

\(\text{5 × 5 = 25}\)

\(\text{5(2x + 5) = 10x + 25}\), so we have factorised correctly.

Example two

Factorise \(\text{3x}^{2} - \text{5x}\)

The HCF of \(\text{3x}^{2}\) and \(\text{5x}\) is \(\text{x}\), so take \(\text{x}\) outside the bracket.

\(\text{3x}^{2} ÷ \text{x = 3x}\)

\(\text{5x} ÷ \text{x = 5}\)

So we have \(\text{x(3x} - \text{5)}\).

Question

Factorise \(\text{6x – 12}\)

Question

Factorise \(\text{10x + x}^{2}\)

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Factorising - Intermediate & Higher tier – WJECOne common factor