Cambridge International Mathematics

Algebra (Expansion and factorisation) (Chapter 1) 33

i ¡2(x+4) j ¡3(2x¡1) k x(x+3) l 2 x(x¡5)

m ¡3(x+2) n ¡4(x¡3) op¡2(x¡y)

q a(a+b) r ¡a(a¡b) s x(2x¡1) t 2 x(x^2 ¡x¡2)

2 Expand and simplify:

a 1+2(x+2) b 13 ¡4(x+3) c 3(x¡2) + 5

d 4(3¡x)¡ 10 e x(x¡1) +x f 2 x(3¡x)+x^2

g 2 a(b¡a)+3a^2 h 4 x¡ 3 x(x¡1) i 7 x^2 ¡ 5 x(x+2)

3 Expand and simplify:

a 3(x¡4) + 2(5 +x) b 2 a+(a¡ 2 b) c 2 a¡(a¡ 2 b)

d 3(y+ 1) + 6(2¡y) e 2(y¡3)¡4(2y+1) f 3 x¡4(2¡ 3 x)

g 2(b¡a)+3(a+b) h x(x+4)+2(x¡3) i x(x+4)¡2(x¡3)

j x^2 +x(x¡1) k ¡x^2 ¡x(x¡2) l x(x+y)¡y(x+y)

m ¡4(x¡2)¡(3¡x) n 5(2x¡1)¡(2x+3) o 4 x(x¡3)¡ 2 x(5¡x)

Consider the product (a+b)(c+d).

It has twofactors, (a+b) and (c+d).

We can evaluate this product by using the distributive law several times.

(a+b)(c+d)=a(c+d)+b(c+d)

=ac+ad+bc+bd

So, (a+b)(c+d)=ac+ad+bc+bd

The final result contains four terms:

ac is the product of the First terms of each bracket.

ad is the product of the Outer terms of each bracket.

bc is the product of the Inner terms of each bracket.

bd is the product of the Last terms of each bracket.

Example 3 Self Tutor

Expand and simplify: (x+ 3)(x+2):

(x+ 3)(x+2)

=x£x+x£2+3£x+3£ 2

=x^2 +2x+3x+6

=x^2 +5x+6

B THE PRODUCT(a+b)(c+d) [2.7]

This is sometimes

called theFOILrule.

In practice we do not

include the second line

of these examples.

¡(7¡ 2 x)

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Y:\HAESE\IGCSE01\IG01_01\033IGCSE01_01.CDR Tuesday, 7 October 2008 12:46:18 PM PETER