Cambridge International Mathematics

Algebra (Expansion and factorisation) (Chapter 1) 45

We know theexpansionof (a+b)(a¡b) is a^2 ¡b^2.

Thus, thefactorisationof a^2 ¡b^2 is (a+b)(a¡b):

a^2 ¡b^2 =(a+b)(a¡b)

In contrast, thesumof two squares does not factorise into two

real linear factors.

Example 22 Self Tutor

Use the rule a^2 ¡b^2 =(a+b)(a¡b) to factorise fully:

a 9 ¡x^2 b 4 x^2 ¡ 25

a 9 ¡x^2

=3^2 ¡x^2 fdifference of squaresg

=(3+x)(3¡x)

b 4 x^2 ¡ 25

=(2x)^2 ¡ 52 fdifference of squaresg

=(2x+ 5)(2x¡5)

Example 23 Self Tutor

Fully factorise: a 2 x^2 ¡ 8 b ¡ 3 x^2 +48

a 2 x^2 ¡ 8

=2(x^2 ¡4)

=2(x^2 ¡ 22 )

=2(x+ 2)(x¡2)

fHCF is 2 g

fdifference of squaresg

b ¡ 3 x^2 +48

=¡3(x^2 ¡16)

=¡3(x^2 ¡ 42 )

=¡3(x+ 4)(x¡4)

fHCF is¡ 3 g

fdifference of squaresg

We notice that x^2 ¡ 9 is the difference of two squares and therefore we can factorise it using

a^2 ¡b^2 =(a+b)(a¡b).

Even though 7 is not a perfect square, we can still factorise x^2 ¡ 7 by writing 7=(

p

7)^2 :

So, x^2 ¡7=x^2 ¡(

p

7)^2 =(x+

p

7)(x¡

p

7):

We say that x+

p

7 and x¡

p

7 are thelinear factorsof x^2 ¡ 7.

[2.8]

H DIFFERENCE OF TWO SQUARES

FACTORISATION

The between

and is

which is the difference

of two squares.

difference

abab^2222 ¡¡¡

Always look to

remove a common

factor first.

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Y:\HAESE\IGCSE01\IG01_01\045IGCSE01_01.CDR Wednesday, 10 September 2008 2:07:46 PM PETER