year 8 maths

1H
Factorising expressions
37
Example 15
Factorising expressions
Factorise the following expressions.
a 6x + 15 b 12a + 18ab c 21x − 14y
d −10a − 20 e x 2 + 4x
SOLUTION
EXPLANATION
a 6x + 15 = 3(2x + 5) HCF of 6x and 15 is 3 . 6x ÷ 3 = 2x and 15 ÷ 3 = 5 .
b 12a + 18ab = 6a(2 + 3b) HCF of 12a and 18ab is 6a . 12a ÷ 6a = 2 and 18ab ÷ 6a = 3b .
c 21x − 14y = 7(3x − 2y) HCF of 21x and 14y is 7 . 21x ÷ 7 = 3x and 14y ÷ 7 = 2y .
The subtraction sign is included, as in the original expression.
d −10a − 20 = −10(a + 2) HCF of −10a and −20 is 10 but because both terms are
negative we bring −10 out the front.
−10a ÷ −10 = a and −20 ÷ −10 = 2 .
e x 2 + 4x = x(x + 4) HCF of x 2 and 4x is x . x 2 ÷ x = x and 4x ÷ x = 4 .
Exercise 1H
UNDERSTANDING AND FLUENCY
1–5, 6–7(½) 4(½), 5, 6–9(½)
7, 8–9(½)
1 List the factors of:
a 20 b 12 c 15 d 27
2 The factors of 14 are 1, 2, 7 and 14 . The factors of 26 are 1, 2, 13 and 26 . What is the highest
factor that 14 and 26 have in common?
3 Find the highest common factor of the following pairs of numbers.
a 12 and 18 b 15 and 25 c 40 and 60 d 24 and 10
4 Fill in the blanks to make these expansions correct.
a 3(4x + 1) = □ x + 3 b 5(7 − 2x) = □ − 10x
c 6(2 + 5y) = □ + □ y d 7(2a − 3b) = □ − □
e 3(2a + □ ) = 6a + 21 f 4( □ − 2y) = 12 − 8y
g 7(□ + □ ) = 14 + 7q h □ (2x + 3y) = 8x + 12y
5 Verify that 5x + 15 and 5(x + 3) are equivalent by copying and completing the table below.
x 2 7 4 0 −3 −6
5x + 15
5(x + 3)
Example 14
6 Find the highest common factor (HCF) of the following pairs of terms.
a 15 and 10x b 20a and 12b c 27a and 9b
d 7xy and 14x e −2yz and 4xy f 11xy and −33xy
g 8qr and −4r h −3a and 6a 2 i 14p and 25pq
Cambridge Maths NSW
Stage 4 Year 8 Second edition
ISBN 978-1-108-46627-1 © Palmer et al. 2018
Cambridge University Press
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