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9. y 12. (a) When x = 1
3(1) – 6(1) – 3(1) + 6 = 0
3
2
x – x – 2
2
3
2
x – 1 x – 2x – x + 2
3
2
x – x
2
15 –x – x
2
–x + x
–2x + 2
x
0.25 3 5 –2x + 2
y = |3x – 6x – 3x + 6|
2
3
= |3(x − 2x − x + 2)|
2
3
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= |3(x − 1)(x − x − 2)|
2
= |3(x − 1)(x + 1)(x − 2)|
[3]
y = x – 7x + 10
2
10. (a) y y = (x – 5)(x – 2)
y
10
x
1 2 4
–24 6
[3]
x
(b) y –1 1 2 5
[8]
(b) 3(x – 1)(x + 1)(x – 2) = (x – 5)(x – 2)
3x – 3 = x – 5
2
24 3x – x + 2 = 0 (rejected)
2
x 3(x – 1)(x + 1)(x – 2) = –(x – 5)(x – 2)
1 2 4
3x – 3 = –x + 5
2
3x + x – 8 = 0
2
x = 1.47
y = (1.475) – 7(1.475) + 10
2
[2] y = 1.85
or
11. y x = –1.81
y = (–1.808) – 7(–1.808) + 10
2
y = 25.92
Intersection points
(1.47, 1.85), (–1.81, 25.92) and (2, 0)
[4]
13. y = –(x + 1)(x – 1)(x – 2)
y = –(x – 1)(x – 2)
2
x
–5 0 5 a = –1, b = –1, c = –2
[3]
[3]
14. 2 = a(2)(–2)(–3)
a = 1
6
b = 2
c = –2
d = –3
Cambridge IGCSE
TM
172 Ace Your Additional Mathematics
Answers Add Math.indd 172 14/03/2022 12:29 PM

