Understanding Engineering Mathematics

When factorising more complicated polynomials it pays to remember that this is not
always possible. For example, we cannot factorisex^2 +1 in terms of real factors, and it
must therefore be left as it is. In any event, factorisation is rarely easy, and by far the most
potent weapon in factorising is a high degree of skill in multiplying out brackets. The
result for the difference of two squares, and the square of a linear factor (Section 2.2.1)
are particularly useful here. In the latter case try to learn to recognise that, provided the
coefficient ofx^2 is 1, the constant term in(x+a)^2 =x^2 + 2 ax+a^2 (i.e. independent of
x) is the square of half of the coefficient of the linear term inx.
Expressions can sometimes be factorised by looking for like terms and combining them,
or for terms with common factors.

Example
2 ac+ad− 6 bc− 3 bd

‘By inspection’ we notice thatamay be taken out of the first two terms to givea( 2 c+d).
We then look at the last two terms to see if they conceal a 2c+d– sure enough they may
be written− 6 bc− 3 bd=− 3 b( 2 c+d).
So we have

2 ac+ad− 6 bc− 3 bd=a( 2 c+d)− 3 b( 2 c+d)

and taking out the common factor 2c+dfinally gives the factorised form

(a− 3 b)( 2 c+d)

Note that this required some inspired guesswork and trial and error.

Solution to review question 2.1.3

A.(i) To factorise 3x^2 + 6 xexamine each term and find factors common

to them. In this case both terms contain anxand a 3, from which

3 x^2 + 6 x=( 3 x)x+( 3 x)( 2 )

= 3 x(x+ 2 )

This done you could, for practise, check the result by expanding

out again, to go from right to left.

(ii) Know your difference of two squares backwards!

u^2 − 16 =u^2 − 42 =(u− 4 )(u+ 4 )

B.(i) In factorising something likex^2 −x−2 most of us use a sort of

inspired trial and error. And the better you are at multiplying pairs

of brackets, the easier the trial will be. We look at the−2. This can

only come from multiplying something like(x± 1 )and(x∓ 2 )

together. Whichever of these we choose must combine to give us

the−x. All you have to do therefore is multiply such pairs of

brackets and check which gives the right result. If your expansion