Cambridge International Mathematics

52 Algebra (Expansion and factorisation) (Chapter 1)

In general, if we start with a quadratic trinomial we will need a method to work out how to do the

splitting.

Consider the expansion in greater detail:

(2x+ 3)(4x+5)=2£ 4 £x^2 +[2£5+3£4]x+3£5=8x^2 +22x+15

The four numbers 2 , 3 , 4 and 5 are present in themiddle term, and also in thefirstandlastterms

combined.

As 2 £ 5 and 3 £ 4 are factors of 2 £ 3 £ 4 £5 = 120, this gives us the method for performing

the splitting.

Step 1: Multiply the coefficient ofx^2 and the constant

term.

Step 2: Look for the factors of this number which add

to give the coefficient of the middle term.

Step 3: These numbers are the coefficients of the split

terms.

In our case, 8 £15 = 120:

What factors of 120 add to give us 22?

The answer is 10 and 12 :

So, the split is 10 x+12x.

Consider another example, 6 x^2 +17x+12.

The product of the coefficient ofx^2 and the constant term is 6 £12 = 72:

We now need two factors of 72 whosesumis 17. These numbers are 8 and 9 :

So, 6 x^2 +17x+12

=6|x^2 {z+8x} +9|x{z+12} f 17 xhas been split into 8 xand 9 xg

=2x(3x+ 4) + 3(3x+4)

=(3x+ 4)(2x+3)

or 6 x^2 +17x+12

=6|x^2 {z+9x} +8|x{z+12}

=3x(2x+ 3) + 4(2x+3)

=(2x+ 3)(3x+4)

What to do:

1 For the following quadratics, copy and complete the table below:

quadratic product sum ‘split’

Example 10 x^2 +29x+21 210 29 14 x+15x

a 2 x^2 +11x+12

b 3 x^2 +14x+8

c 4 x^2 +16x+15

d 6 x^2 ¡ 5 x¡ 6

e 4 x^2 ¡ 13 x+3

f 6 x^2 ¡ 17 x+5

2 Use your tabled results to factorise each of the quadratics in 1.

When splitting the

middle term it does

not matter the order

in which you list the

two new terms.

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Y:\HAESE\IGCSE01\IG01_01\052IGCSE01_01.CDR Wednesday, 10 September 2008 12:19:01 PM PETER