Everything Maths Grade 12

5.3 CHAPTER 5. FACTORISING CUBIC POLYNOMIALS

Since g(−4) = 0, y + 4 is a factor of g(y) = 5y^4 + 16y^3 − 15 y^2 + 8y + 16.

5.3 Factorisation of Cubic Polynomials

EMCAR

A cubic polynomial is apolynomial of the form

ax^3 +bx^2 +cx +d

where a is nonzero. Wehave seen in Grade 10that the sum and difference of cubes is factorised as

follows:

(x +y)(x^2 −xy +y^2 ) = x^3 +y^3

and

(x−y)(x^2 +xy +y^2 ) = x^3 −y^3

We also saw that the quadratic factor does not have real roots.

There are many methods of factorising a cubic polynomial. The generalmethod is similar to thatused

to factorise quadratic equations. If you have a cubic polynomial of the form:

f (x) = ax^3 +bx^2 +cx +d

then in an ideal world you would get factors of the form:

(Ax +B)(Cx +D)(Ex +F ). (5.1)

But sometimes you willget factors of the form:

(Ax +B)(Cx^2 +Ex +D)

We will deal with simplest case first. When a = 1, then A = C = E = 1, and you only have to

determine B, D and F. For example, find thefactors of:

x^3 − 2 x^2 − 5 x + 6.

In this case we have

a = 1

b =− 2

c =− 5

d = 6

The factors will have thegeneral form shown in (5.1), with A = C = E = 1. We can then use values

for a, b, c and d to determine values for B, D and F. We can re-write (5.1) with A = C = E = 1 as:

(x +B)(x +D)(x +F ).

If we multiply this out we get:

(x +B)(x +D)(x +F ) = (x +B)(x^2 +Dx +Fx +DF )

= x^3 +Dx^2 +Fx^2 +Bx^2 +DFx +BDx +BFx +BDF

= x^3 + (D +F +B)x^2 + (DF +BD +BF )x +BDF

We can therefore write:

b =−2 = D +F +B (5.2)

c =−5 = DF +BD +BF (5.3)

d = 6 = BDF. (5.4)