Understanding Engineering Mathematics

This is fine, but it is much quicker if you notice that pairing off one

of the(x− 1 )factors with the(x+ 1 )givesx^2 −1 and:

(x− 1 )^2 (x+ 1 )=(x− 1 )(x^2 − 1 )

=x^3 −x^2 −x+ 1

is obtained directly.

(iv) No tricks here, just a careful plod:

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

= 2 x^3 +x^2 +x− 1

Note that in each of the above examples, we can always check our results

by substituting ‘obvious’ values ofxin the result.x=0 is often a good

start, whereasx=±1 can be used in (iii), the point being that your calcu-

lated results must vanish for these values.x=^12 will do a similar (but

messy) job in (iv).

2.2.3 Factorisation of polynomials by inspection

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As mentioned in Section 2.2.1 it is sometimes necessary to reverse the multiplication
of polynomials or brackets to split an expression intofactorsof a particular type. This
process is calledfactorising. It is useful in solving algebraic equations, or determining
their properties. Usually we try to factorise into ‘linear’ factors of the formx−a.

Example

You can confirm thatx^2 + 2 x+ 1 =(x+ 1 )^2. This tells us that the quadratic equation
x^2 + 2 x+ 1 =0 has two equal rootsx=−1. It also makes clear thatx^2 + 2 x+1is
always positive whatever the value ofx.
In general, factorising can be difficult. It pays to take it slowly and build up confidence
with simple expressions.

Example

To factorisex^2 + 2 xnote thatxis a common factor of each of the terms in the polynomial.
We can therefore ‘take it out’ and write

x^2 + 2 x=x(x+ 2 )

The approach in general is to inspect each term of an expression and check whether there
are factors common to each. All we are really doing is reversing thedistributive rule
stated in Section 2.2.1.

Example
8 x^2 − 2 x^4 = 4 ( 2 x^2 )−( 2 x^2 )x^2
= 2 x^2 ( 4 −x^2 )
= 2 x^2 ( 2 −x)( 2 +x)

by using the difference of two squares (see Section 2.2.1).