Everything Maths Grade 12

CHAPTER 5. FACTORISING CUBIC POLYNOMIALS 5.3

This is a set of three equations in three unknowns. However, we know that B, D and F are factors of
6 because BDF = 6. Therefore we can use atrial and error method tofind B, D and F.
This can become a verytedious method, therefore the Factor Theorem can be used to find thefactors
of cubic polynomials.

Example 3: Factorisation of CubicPolynomials

QUESTION

Factorise f (x) = x^3 +x^2 − 9 x− 9 into three linear factors.

SOLUTION

Step 1 : By trial and error usingthe factor theorem to find a factor

Try

f (1) = (1)^3 + (1)^2 − 9(1)− 9 = 1 + 1− 9 − 9 =− 16

Therefore (x− 1) is not a factor

Try

f (−1) = (−1)^3 + (−1)^2 9(−1)9 = 1 + 1 + 99 = 0

Thus (x + 1) is a factor, because f (−1) = 0.

Now divide f (x) by (x + 1) using division by inspection:

Write x^3 +x^2 − 9 x− 9 = (x + 1)( )

The first term in the second bracket must bex^2 to givex^3 if one works backwards.

The last term in the second bracket must be− 9 because +1×−9 =− 9.

So we have x^3 +x^2 − 9 x− 9 = (x + 1)(x^2 +?x− 9).

Now, we must find the coefficient of the middleterm (x).

(+1)(x^2 ) gives the x^2 in the original polynomial. So, the coefficient ofthe x-term

must be 0.

So f (x) = (x + 1)(x^2 − 9).

Step 2 : Factorise fully

x^2 − 9 can be further factorised to (x− 3)(x + 3),

and we are now left with f (x) = (x + 1)(x− 3)(x + 3)

In general, to factorise acubic polynomial, you find one factor by trial anderror. Use the factor theorem
to confirm that the guessis a root. Then divide thecubic polynomial by thefactor to obtain a quadratic.
Once you have the quadratic, you can apply thestandard methods to factorise the quadratic.

For example the factorsof x^3 − 2 x^2 − 5 x + 6 can be found as follows: There are three factors which
we can write as
(x−a)(x−b)(x−c).