Cambridge International AS and A Level Mathematics Pure Mathematics 1

Algebra

P1^

1

EXAMPLE 1.22 Factorise x^2 + 7 x + 12.

SOLUTION

Splitting the middle term, 7x, as 4x + 3 x you have

x^2 + 7 x + 12 = x^2 + 4 x + 3 x + 12

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

= (x + 3)(x + 4).

How do you know to split the middle term, 7x, into 4x + 3 x, rather than say

5 x + 2 x or 9x − 2 x?

The numbers 4 and 3 can be added to give 7 (the middle coefficient) and

multiplied to give 12 (the constant term), so these are the numbers chosen.

x^2 + 7 x + 12

EXAMPLE 1.23 Factorise x^2 − 2 x − 24.

SOLUTION

First you look for two numbers that can be added to give −2 and multiplied to

give –24:

− 6 + 4 = − 2 − 6 × (+4) = −24.

The numbers are –6 and +4 and so the middle term, –2x, is split into –6x + 4 x.

x^2 – 2x – 24 = x^2 − 6 x + 4 x − 24

= x(x − 6) + 4(x − 6)

= (x + 4)(x − 6).

x^2

4 x

3 x

12

x 3

x

4

Figure 1.5

The coefficient of x is 7. The constant term is 12.

4 + 3 = 7^4 × 3 = 12