Given a function $f(x)$ , a factor is a number $a$ such that $f(a) = 0$ . A factor is represented as $(x-a)$ . The factor theorem is one of the easiest ways of finding out if a term is a linear factor of a polynomial.

The factor theorem

The factor theorem states that, given a polynomial function $f(x)$ , then:

if $(x-a)$ is a factor of $f(x)$ , then $f(a)=0$ ; and
if $f(a)=0$ , then $(x-a)$ is a factor of $f(x)$ .

Using the factor theorem to factorise a cubic function

Procedure

1.	Substitute values $a$ into the cubic $f(x)$ until you find a value that gives as a result $f(a) = 0$ .
2.	Divide your function $f(x)$ by $(x-a)$ . This can be done with long division for example.
3.	Write your cubic in the following form: $f(x) = (x-a)(Ax^2 + Bx +C)$ where $A$ , $B$ and $C$ are constants.
4.	Factorise the quadratic term $(Ax^2 + Bx + C)$ if possible, in order to express $f(x)$ a function of linear factors.

Example 1

Use the factor theorem to prove that $(x-4)$ is a factor of $f(x)= -3x^3+13x^2 -6x+8$ , and fully factorise $f(x)$ .

To prove that $(x-4)$ is a factor of $f(x)=-3x^3+13x^2 -6x+8$ , you have to substitute $x=4$ into your function.

$f(4) = -3(4)^3 + 13 (4)^2 - 6 \times 4 + 8 = -3 \times 64 + 13 \times 16 - 24 +8=0$ 

Since $f(4) =0$ , you have proven that $(x-4)$ is a factor of $f(x)=-3x^3+13x^2 -6x+8$ .

Divide your polynomial by the factor, and see if the remainder is $0$ . Also obtain the quadratic counterpart.

$\begin{array}{r}-3x^2 +x -2 \hspace{0mm} \\x-4{\overline{\smash{\big)}\, -3x^3 +13x^2 -6x + 8}}\\ \hspace{35mm}- 3x^3 + 12x^2\hspace{14mm}\\\overline{\smash{}\,\hspace{15mm} x^2 -6x+8}\\ \hspace{24mm}x^2 -4x \hspace{6mm}\\\overline{\smash{}\,\hspace{3mm} \ -2x + 8 }\\-2x+8 \\\overline{\space\space\space\space\space\space\space\space\space\space\space\space0}\end{array}$

Now you can write $f(x)$ as

$f(x) = (x-4) (-3x^2 + x +2)$

Factorise the term $(-3x^2 +x -2)$ to present $f(x)$ as a product of linear factors.

$\underline{f(x) = -(x-4)(x-1)\left(3x+2\right)}$ 

Example 2

Show that $(x-5)$ is a factor of $f(x)=x^3 -7x^2 +2x + 40$ , and fully factorise $f(x)$ .

Since you are trying to prove that $(x-5)$ is a factor of $f(x)=x^3 -7x^2 +2x + 40$ , you substitute $x=5$ into your function.

$f(5)=5^3 -7\times 5^2 +2\times 5 + 40 = 125 - 175 + 10 +40=0$ 

Because $f(5)=0$ , you have proven that $(x-5)$ is a factor of $f(x)=x^3 -7x^2 +2x + 40$ .

To find the quadratic factor, you divide your polynomial by the newfound linear factor. This also acts as a double-check, since if the remainder is $0$ , then your linear factor is indeed a factor of $f(x)$ .

$\begin{array}{r}x^2 -2x -8 \hspace{0mm} \\x-5{\overline{\smash{\big)}\, x^3 -7x^2 +2x +40}}\\\hspace{0mm}x^3 - 5x^2\hspace{16mm}\\\overline{\smash{}\,\hspace{5mm} -2x^2 +2x+40}\\\hspace{25mm}-2x^2 +10x \hspace{6mm}\\\overline{\smash{}\,\hspace{10mm} -8x +40}\\\hspace{15mm}-8x+40 \\\overline{\smash{}\,\hspace{10mm} \ 0 }\end{array}$

So $(x-5)$ is indeed a factor of $f(x) = x^3 -7x^2 + 2x + 40$ .

Hence,

$f(x) = (x-5)(x^2-2x-8)$

Now factorise, if possible, the quadratic term $(x^2 - 2x -8)$ to express $f(x)$ as a product of linear factors. However, you will see that this term does not have any real roots, so you leave it with a quadratic term.

Therefore $\underline{f(x) = (x-5)(x^2 -2x +8)}$ is fully factorised.

Learn with Basics

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Unit 1

Quadratic graphs

Unit 2

Factorising quadratics

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Unit 3