Factorising cubics: the remainder and factor theorems

Remainder theorem. If a polynomial $f(x)$ is divided by $x-a$, the remainder is $f(a)$.

If you divide by $ax-b$, the remainder is $f\left(\frac{b}{a}\right)$, because $ax-b=0$ when $x=\frac{b}{a}$.

Example. Find the remainder when $f(x)=x^3-4x^2+2x+7$ is divided by $x+2$.

Here $x+2=0$ when $x=-2$, so the remainder is

$f(-2)=-8-16-4+7=-21$

Watch the sign: dividing by $x+2$ means you substitute $x=-2$.

Factor theorem. $x-a$ is a factor of $f(x)$ if and only if $f(a)=0$. In other words, a factor is a divisor that leaves a remainder of 0.

To find a first factor of a cubic with integer coefficients, test values of $x$ that divide the constant term. If the leading coefficient is not 1, also try fractions of the form $\frac{\text{factor of the constant}}{\text{factor of the leading coefficient}}$.

Example. For $f(x)=x^3+2x^2-5x-6$, the constant term is $-6$, so try $\pm 1$, $\pm 2$, $\pm 3$, $\pm 6$.

$f(1)=1+2-5-6=-8\ne 0$

$f(-1)=-1+2+5-6=0$

So $x+1$ is a factor.

Once you have one factor $x-a$, divide it out to get a quadratic, then factorise the quadratic if possible. The IEB accepts any valid method.

Method 1: compare coefficients. Write $x^3+2x^2-5x-6=(x+1)(x^2+bx-6)$. The first and last terms of the quadratic are chosen so the $x^3$ and constant terms match. Expanding, the $x^2$ term is $(b+1)x^2$, so $b+1=2$ and $b=1$:

$x^3+2x^2-5x-6=(x+1)(x^2+x-6)=(x+1)(x+3)(x-2)$

Method 2: synthetic division. Divide $2x^3-x^2-13x-6$ by $x-3$ (since $f(3)=54-9-39-6=0$). Write the coefficients $2$, $-1$, $-13$, $-6$ and work with 3:

• bring down 2
• $2\times 3=6$; $-1+6=5$
• $5\times 3=15$; $-13+15=2$
• $2\times 3=6$; $-6+6=0$ (the remainder)

The quotient is $2x^2+5x+2$, so

$2x^3-x^2-13x-6=(x-3)(2x^2+5x+2)=(x-3)(2x+1)(x+2)$

If a power of $x$ is missing, write a 0 for its coefficient. For $x^3-7x-6$ the coefficients are $1$, $0$, $-7$, $-6$.

Grouping. Sometimes the terms group neatly: $x^3-2x^2+3x-6=x^2(x-2)+3(x-2)=(x-2)(x^2+3)$.

"Factorise fully" means every factor that can be factorised over the real numbers has been. A quadratic factor with a negative discriminant, such as $x^2+3$, cannot be factorised further.

To solve $f(x)=0$ for a cubic, factorise fully and set each factor equal to 0.

Example. Solve $x^3-3x^2-2x+4=0$.

$f(1)=1-3-2+4=0$, so $x-1$ is a factor and

$x^3-3x^2-2x+4=(x-1)(x^2-2x-4)$

The quadratic does not factorise, so use the quadratic formula:

$x=\frac{2\pm \sqrt{4+16}}{2}=1\pm \sqrt{5}$

The solutions are $x=1$, $x=1+\sqrt{5}\approx 3,24$ and $x=1-\sqrt{5}\approx -1,24$.

A cubic always has at least one real root. If the quadratic factor has a negative discriminant, the cubic has only the one real root from the linear factor.

If you are told that $x-a$ is a factor, write $f(a)=0$. If you are told the remainder on division by $x-a$ is $R$, write $f(a)=R$. Each fact gives one equation, so two unknowns need two facts, which you then solve simultaneously.

Example. $x-2$ is a factor of $f(x)=x^3+k{x}^2-x-6$. Find $k$.

$f(2)=8+4k-2-6=0\Rightarrow 4k=0\Rightarrow k=0$

So $f(x)=x^3-x-6=(x-2)(x^2+2x+3)$, and $x^2+2x+3$ has discriminant $4-12<0$, so this is the full factorisation.