Understanding Engineering Mathematics

Example
p(x)≡x^4 + 2 x^3 −x− 2

The roots must be factors of−2 and so the only integer roots are likely to be±1or±2.
Wefindbytrialthat

p( 1 )=0 giving a factorx− 1

p(− 2 )=0 giving a factorx+ 2

However,−1 and 2 are not roots, so we have probably exhausted all possible real factors
(although there may be repeated roots, see below). To find the quadratic factor remaining
we write

x^4 + 2 x^3 −x− 2 ≡(x− 1 )(x+ 2 )(x^2 +ax+b)

≡(x^2 +x− 2 )(x^2 +ax+b)

and find, on multiplying out the right-hand side and equating coefficients in the resulting
identity

− 2 b=− 2 , sob= 1 (constant term)

b− 2 a=− 1 , soa= 1 (coefficient ofx)

So the factorised form is

p(x)=(x− 1 )(x+ 2 )(x^2 +x+ 1 )

The quadraticx^2 +x+1 does not factor further into real factors – it is said to beirre-
ducible.
It may not always be clear whether or not there are repeated factors in a polynomial.
This does not really matter much since any repeated factors will become apparent from
the remaining quadratic factor for simple cases. There is a result that ifx−ais a repeated
factor of a polynomialp(x)then it is also a factor of the derivativep′(x), but this is rarely
worth using.

Solution to review question 2.1.6

A.From Review Question 2.1.2(iii) we have

x^3 −x^2 −x+ 1 =(x− 1 )^2 (x+ 1 )

(i) The factors are therefore(x− 1 )(twice) andx+1.

(ii) The roots are the values which make the polynomial vanish, which

from the factors arex=1 (twice) andx=−1.

Note that a factor is an expression containingx while a root is a

particular value ofx.

The solutions of the equation

x^3 −x^2 −x+ 1 = 0