GCSE Maths revision · Higher: number and algebra

Harder expanding and factorising

Higher only

The idea

Expand three brackets by multiplying two first, then multiplying the result by the third. To factorise $ax^2 + bx + c$ , find two numbers that multiply to $ac$ and add to $b$ , then split the middle term.

For $6x^2 + 7x - 3$ : $ac = -18$ , and $9$ and $-2$ add to $7$ .
Split: $6x^2 + 9x - 2x - 3$ .
Factorise in pairs: $3x(2x + 3) - 1(2x + 3)$ becomes $(3x - 1)(2x + 3)$ .

Watch out Take out any common factor first, and expand your answer to check.

See it worked

Worked exampleNon-calculator4 marks

Factorise each expression.

(a) Factorise $2x^2 + 7x + 3$

(b) Factorise $6x^2 - x - 2$

(a) $2 \times 3 = 6$ : find two numbers that multiply to 6 and add to 7: $6$ and $1$ .
Split the middle term and group. $2x^2 + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3)$
(b) $6 \times (-2) = -12$ : numbers that multiply to $-12$ and add to $-1$ are $-4$ and $3$ .
$6x^2 - 4x + 3x - 2 = 2x(3x - 2) + 1(3x - 2) = (3x - 2)(2x + 1)$

Answer (a) $(2x + 1)(x + 3)$ ; (b) $(3x - 2)(2x + 1)$

Your turn

Question 1Non-calculator2 marks

Factorise fully $6x^{2}$ + x $-$ 12.

Question 2Non-calculator4 marks

Consider $10x^2 - 13x - 3$ .

(a) Factorise $10x^2 - 13x - 3$ .

(b) Hence solve $10x^2 - 13x - 3 = 0$ .

Question 3Non-calculator4 marks

Consider $6x^2+38x+56$ .

(a) Factorise fully.

(b) Solve the equation obtained by setting the expression equal to zero.

In past papers

62 official questions in 54 papers since 2017. The latest:

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