Understanding Engineering Mathematics

of brackets is good, then you’ll be able to do this mentally:

(x− 1 )(x+ 2 )=x^2 +x− 2 ×

(x+ 1 )(x− 2 )=x^2 −x− 2

√

So the factors are

(x+ 1 )(x− 2 )

If you want a more formal approach, note that

(x+a)(x+b)=x^2 +(a+b)x+ab

So, to factorisex^2 −x−2 we need to findaandbsuch that

ab=−2anda+b=−1, which is then done ‘by inspection’

(i.e. trial and error!) to give the result obtained above.

(ii) There is a routine device for factorising polynomials such as

x^3 − 2 x^2 −x+2 or higher degree, based on the factor theorem

(see Section 2.2.6). This only works if the roots of the polynomial

are easy to spot. This example shows what can be done by looking

at the terms and picking factors out directly. In this case a couple

of elementary factorisations reveals that

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

We continue by taking out the(x− 2 )factor now common to the

two terms

=(x^2 − 1 )(x− 2 )

and finish by factorising thex^2 − 1

=(x− 1 )(x+ 1 )(x− 2 )

This example nicely illustrates how one often has to piece together

elementary results and ideas to solve problems.

(iii) For 3x^2 + 5 x−2 we have to deal with the 3, the coefficient ofx.

Because of this we guess that the factors are going to be some-

thing like( 3 x+a)(x+b), withab=−2, anda+ 3 b=5. Trial

and error (which can be systematised, but is hardly worth the

effort) soon reveals thatb=2anda=−1 works and we can

then quickly check that( 3 x− 1 )(x+ 2 )= 3 x^2 + 5 x−2.

Note that in these solutions we don’t want to give the impression that
factorisation is an haphazard, hit and miss process – there are system-
atic routines, including symbolic computer algebra packages – we simply
want to indicate what can be achieved with a good facility in simple
algebra.