Expanding and factorising

Foundation and Higher

The idea

Expanding multiplies every term inside the bracket: $3(2x - 5) = 6x - 15$ . Factorising is the reverse: take out the highest common factor, $6x - 15 = 3(2x - 5)$ .

Multiply the term outside by each term inside, signs included.
To factorise, find the HCF of the numbers and of the letters.
Check by expanding your answer.

Watch out $-2(x - 4) = -2x + 8$ : the minus multiplies every term.

See it worked

Worked exampleNon-calculator4 marks

Answer each part.

(a) Expand $5(x + 4)$

(b) Expand and simplify $3(2y - 1) + 4y$

(c) Factorise $6a + 15$

(a) Multiply each term in the bracket by 5: $5 \times x + 5 \times 4 = 5x + 20$ .
(b) Expand, then collect like terms. $6y - 3 + 4y = 10y - 3$
(c) The highest common factor of 6 and 15 is 3: $6a + 15 = 3(2a + 5)$ .

Answer (a) $5x + 20$ ; (b) $10y - 3$ ; (c) $3(2a + 5)$

Your turn

Question 1Calculator allowed2 marks

Expand and simplify 9 $-$ 3(2x $-$ 5).

Question 2Non-calculator4 marks

A pupil is simplifying $6-3(2x-2)$ .

(a) Expand and simplify the expression.

(b) Find x when this expression is zero.

Question 3Non-calculator5 marks

The perimeter of a triangle is $(10x + 5)$ cm. Two of its sides are $2(x + 3)$ cm and $3(x - 1)$ cm.

(a) Find an expression for the third side. Simplify your answer.

(b) Factorise the perimeter expression fully.

In past papers

126 official questions in 106 papers since 2017. The latest:

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