Factorising Expressions

Factorising is reverse expanding

When you expand, you multiply a bracket out into a sum: 3(2x + 3) becomes 6x + 9. Factorising runs that process backwards. You start with a sum like 6x + 9 and rewrite it as a product with a bracket: 3(2x + 3).

If you are confident with Expanding Brackets, factorising is the same skill in reverse — and expanding is exactly how you'll check your work.

Why bother? Factorising reveals hidden structure. It is the key to simplifying algebraic fractions, solving quadratic equations, and rearranging formulas. A factorised expression often shows you a shortcut the expanded form hides.

The highest common factor (HCF)

To factorise, you look for what every term has in common and pull it out to the front. The most you can pull out is the highest common factor (HCF) — the largest number and the most variables that divide into every term.

Consider 6x + 9:

• The coefficients are 6 and 9. The largest number dividing both is 3.
• Only the first term has an x, so no variable is common.
• The HCF is therefore 3.

If you need to brush up on finding common factors of numbers, see Factors and Multiples.

The three-step method

1. Find the HCF of all the terms (numbers and variables).
2. Write the HCF outside a bracket.
3. Divide each original term by the HCF to find what goes inside.

Then check by expanding.

Worked example 1 — numbers only

Factorise 6x + 9.

1. HCF of 6 and 9 is 3. (No variable is shared.)
2. Write the 3 outside: 3( … ).
3. Divide each term by 3: 6x ÷ 3 = 2x, and 9 ÷ 3 = 3.
4. Fill the bracket: 3(2x + 3).

Check by expanding: 3 × 2x = 6x and 3 × 3 = 9, giving 6x + 9. ✓ It matches.

Worked example 2 — fully factorising

Factorise 4x + 8.

1. HCF of 4 and 8 is 4.
2. Outside: 4( … ).
3. Divide: 4x ÷ 4 = x, and 8 ÷ 4 = 2.
4. Result: 4(x + 2).

A common mistake is to stop at 2(2x + 4). That is partly factorised, but 2x + 4 still has a common factor of 2, so it is not fully factorised. Always take out the highest common factor in one go.

Worked example 3 — a common variable

Factorise x² + 5x.

Here the shared factor is a variable, not a number.

1. x² means x · x, and 5x means 5 · x. Both terms contain at least one x.
2. The HCF is x (the coefficients 1 and 5 share only 1).
3. Outside: x( … ).
4. Divide: x² ÷ x = x, and 5x ÷ x = 5.
5. Result: x(x + 5).

Check: x × x = x² and x × 5 = 5x, giving x² + 5x. ✓

Worked example 4 — number and variable together

Factorise 6x² − 4x.

Now both a number and a variable are common.

1. Coefficients 6 and 4 have HCF 2.
2. Both terms contain at least one x (x² and x), so x is common.
3. The full HCF is 2x.
4. Outside: 2x( … ).
5. Divide each term by 2x: 6x² ÷ 2x = 3x, and −4x ÷ 2x = −2. Keep the minus sign.
6. Result: 2x(3x − 2).

Check: 2x × 3x = 6x² and 2x × (−2) = −4x, giving 6x² − 4x. ✓

Worked example 5 — three terms

Factorise 12a + 18ab − 6a.

The method works no matter how many terms there are.

1. Coefficients 12, 18, and 6 have HCF 6.
2. Every term contains an a, so a is common. (Not every term has a b, so b is not common.)
3. Full HCF: 6a.
4. Divide: 12a ÷ 6a = 2; 18ab ÷ 6a = 3b; −6a ÷ 6a = −1.
5. Result: 6a(2 + 3b − 1), which simplifies inside to 6a(1 + 3b).

Tidying the bracket (2 − 1 = 1) is a good habit; always simplify what is left inside.

How to check you've fully factorised

After factorising, ask: does anything inside the bracket still share a common factor? If yes, you didn't take out the HCF. For example, 2(3x + 6) is wrong because 3x + 6 shares a 3; the fully factorised form is 6(x + 2). Expanding to confirm you recover the original is the surest check of all.

Activity — factorise each expression

1. 5x + 10
2. 8x − 12
3. x² + 7x
4. 9x + 6
5. 4x² + 10x
6. 14a − 21ab

Answers: 1) 5(x + 2) 2) 4(2x − 3) 3) x(x + 7) 4) 3(3x + 2) 5) 2x(2x + 5) 6) 7a(2 − 3b)

Why this matters

Factorising is one of the most reused skills in algebra. It lets you cancel common factors in fractions, and it is the foundation for solving quadratic equations, where writing an expression as a product is exactly what makes the solution appear. Mastering the HCF method now pays off across all of higher mathematics.

The whole process is just three moves: find the HCF, place it outside, divide each term to fill the bracket — then expand to check. If anything inside still shares a factor, go back and take more out.