Understanding Engineering Mathematics

This can only be true if eitherx− 1 =0orx− 2 =0, yielding two possible values ofx:

x= 1 , 2

Such solutions of a polynomial equation are called therootsorzerosof the corresponding
polynomial (➤52).
Some polynomials do not have roots that are real numbers – for example there are no
real numbers that satisfy

x^2 + 1 = 0

However, if we allow the possibility ofcomplex roots(complex numbers are considered
in Chapter 12), then there is a famous theorem of algebra (thefundamental theorem of
algebra) which states that a polynomial ofnth degree has exactlynroots – which may
be real, equal or complex.
Polynomials may be added and multiplied to produce other polynomials. In doing so
remember to gather like terms, and take care with signs.

Examples
( 2 x+ 1 )+(x^3 − 2 x^2 − 3 x− 7 )=x^3 − 2 x^2 +( 2 x− 3 x)+( 1 − 7 )
=x^3 − 2 x^2 −x− 6

( 2 x^2 +x− 1 )(x^2 − 2 )= 2 x^2 x^2 + 2 x^2 (− 2 )+xx^2 +x(− 2 )− 1 x^2 − 1 (− 2 )

= 2 x^4 − 4 x^2 +x^3 − 2 x−x^2 + 2

= 2 x^4 +x^3 − 5 x^2 − 2 x+ 2

Solution to review question 2.1.2

(i) We must gather together all terms of the same degree:

x^3 − 2 x+ 1 + 3 (x^4 + 2 x^3 − 4 x^2 −x− 1 )

=x^3 − 2 x+ 1 + 3 x^4 + 6 x^3 − 12 x^2 − 3 x− 3

= 3 x^4 + 7 x^3 − 12 x^2 − 5 x− 2

(ii) Expanding the brackets:

(x− 2 )(x+ 3 )=x(x+ 3 )− 2 (x+ 3 )

=x^2 + 3 x− 2 x− 6

=x^2 +x− 6

(iii) This example illustrates how good facility with elementary results

can save you work. Thus, one way to expand in this case is:

(x− 1 )^2 (x+ 1 )=(x^2 − 2 x+ 1 )(x+ 1 )

=x^3 − 2 x^2 +x+x^2 − 2 x+ 1

=x^3 −x^2 −x+ 1