M8 · Expanding and factorising | Edexcel GCSE Maths

Revise the key ideas

Distribute the term outside a bracket to every term inside: 3(x + 4) = 3x + 12. A minus sign outside changes every sign: −(2x − 5) = −2x + 5.
Worked example: 4(2x − 3) − 2(x + 5) = 8x − 12 − 2x − 10 = 6x − 22. Expand before collecting like terms.
For two brackets, multiply every term in one by every term in the other. (x + 3)(x + 2) = x² + 2x + 3x + 6 = x² + 5x + 6.
Area model for expanding two brackets
Enlarge diagram: Area model for expanding two brackets
Useful identities are (a + b)² = a² + 2ab + b² and (a − b)² = a² − 2ab + b². The middle term is essential.

Factorising reverses expanding. Take out the highest common factor: 12x² + 8x = 4x(3x + 2). Expanding again checks both terms.
To factorise x² + bx + c, find two numbers whose sum is b and product is c. x² + 7x + 12 = (x + 3)(x + 4).
Worked example: x² − x − 12 = (x − 4)(x + 3), because −4 + 3 = −1 and (−4) × 3 = −12.
A difference of two squares is a² − b² = (a − b)(a + b). Thus x² − 25 = (x − 5)(x + 5). A sum of squares does not use this identity.
Factorising an expression does not by itself find x. To solve an equation, first make one side zero, then use the zero-product rule.

For ax² + bx + c with a ≠ 1, find a factorisation that produces both the correct leading coefficient and middle term.
Worked example: 6x² + 7x + 2 = 6x² + 3x + 4x + 2 = 3x(2x + 1) + 2(2x + 1) = (3x + 2)(2x + 1).
For three brackets, expand two, simplify, then multiply the result by the third. (x + 1)(x − 1)(x + 2) = (x² − 1)(x + 2) = x³ + 2x² − x − 2.
Look for a common factor before a quadratic pattern: 2x² − 18 = 2(x² − 9) = 2(x − 3)(x + 3).

Practice questions with explained answers will be added here.

For now, cover the worked answers, try the calculations yourself, then compare each step. Include units and reasons where needed.