Understanding Engineering Mathematics

are, of course, simply the roots of the polynomial on the left-hand side,

by definition, and so the solutions arex=1 (twice) andx=−1.

B. No obvious factors present themselves inp(x)=x^3 + 2 x^2 − 5 x−6.

The best approach here is to use the factor theorem. We know that any

linear factor of the formx−amust be such thatais a factor of−6.

Also, by the factor theorem, ifx−ais a factor of the polynomialp(x),

thenp(a)=0. All this suggests that we look for roots of the polyno-

mial amongst the factors of−6, by substituting these in the polynomial.

Tryingx=1:

p( 1 )= 13 + 2 × 12 − 5 × 1 − 6 =− 8    = 0

So(x− 1 )is not a factor.

Butx=−1gives

(− 1 )^3 + 2 (− 1 )^2 − 5 (− 1 )− 6 = 0

So(x−(− 1 ))=(x+ 1 )is a factor.

Similarly, you can check thatx=−2,x=3 are not roots, butx= 2

andx=−3 are, giving the final factorisation as

(x+ 1 )(x− 2 )(x+ 3 )

NB. By far the most common mistake in this topic is to think that if

x=ais a root thenx+ais the corresponding factor. This error is

particularly easily made ifais negative. Remember, the correct factor

isx−a.

2.2.7 Rational functions

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39 76 ➤

In writing down a polynomial we need only addition (or subtraction) and multiplication.
The next step is to consider algebraicfractions, involving the division of polynomials.
The simplest example is

1

x

This already gives us new problems. Whereas a polynomial inxexists for every value
ofx– that is, plug in a value ofxand you will get out a value for the polynomial – the

above fractiondoes not exist forx= 0. That is, there is no such number as

1

0

. As noted in

Section 1.2.1 it simply does not exist – we say

1

x

is not defined atx= 0. By this we mean

that we are not allowed to putx=0 in this expression. For this reason we should really write

1

x

x   = 0

but the ‘x =0’ is often omitted, so long as it is clearly understood.