Cambridge International Mathematics

48 Algebra (Expansion and factorisation) (Chapter 1)

Some expressions with four terms do not have an overall common factor, but can be factorised by pairing

the four terms.

For example, ab| +{zac}+bd| +{zcd}

=a(b+c)+d(b+c) ffactorising each pair separatelyg

=(b+c)(a+d) fremoving the common factor(b+c)g

Note:

² Many expressions with four terms cannot be factorised using this method.

² Sometimes it is necessary to reorder the terms first.

Example 28 Self Tutor

Factorise: a 3 ab+d+3ad+b b x^2 +2x+5x+10

a 3 ab+d+3ad+b

=3ab+b

| {z }

+3ad+d

| {z }

=b(3a+1)+d(3a+1)

=(3a+1)(b+d)

b x|^2 {z+2x}+5|x{z+10}

=x(x+2)+5(x+2)

=(x+2)(x+5)

Example 29 Self Tutor

Factorise: a x^2 +3x¡ 4 x¡ 12 b x^2 +3x¡x¡ 3

a x|^2 +3{z x}¡| 4 x{z¡ (^12) }
=x(x+3)¡ 4 (x+3)
=(x+3)(x¡4)
b x|^2 {z+3x}¡| x{z¡ (^3) }
=x(x+3)¡(x+3)
=x(x+3)¡1(x+3)
=(x+3)(x¡1)
EXERCISE 1J
1 Factorise:
a 2 a+2+ab+b b 4 d+ac+ad+4c c ab+6+2b+3a
d mn+3p+np+3m e x^2 +3x+7x+21 f x^2 +5x+4x+20
g 2 x^2 +x+6x+3 h 3 x^2 +2x+12x+8 i 20 x^2 +12x+5x+3
2 Factorise:
a x^2 ¡ 4 x+5x¡ 20 b x^2 ¡ 7 x+2x¡ 14 c x^2 ¡ 3 x¡ 2 x+6
d x^2 ¡ 5 x¡ 3 x+15 e x^2 +7x¡ 8 x¡ 56 f 2 x^2 +x¡ 6 x¡ 3
g 3 x^2 +2x¡ 12 x¡ 8 h 4 x^2 ¡ 3 x¡ 8 x+6 i 9 x^2 +2x¡ 9 x¡ 2
J EXPRESSIONS WITH FOUR TERMS [2.8]
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Y:\HAESE\IGCSE01\IG01_01\048IGCSE01_01.CDR Wednesday, 10 September 2008 2:08:09 PM PETER