Fully factorise $x^3-x^2-14x+24$ into linear factors

Question

$$f(x)=x^3-x^2-14x+24$$

I've tried grouping the terms, but it just doesn't work out for me. Any help is appreciated.

Answer 1

We can hope there is a rational root. Real world cubics generally don't have such a root, but school cubics often do.

By the Rational Root Theorem, such a root has to be an integer that divides $24$. Soon we find that $x=2$ works.

Divide the given cubic polynomial by $x-2$. We get a quadratic, and you know how to find roots of quadratics.

Answer 2

The factor theorem states that if $x-a$ is a factor of some polynomial $f(x)$, then $f(a)=0$. The proof of this is very straightforward. Try it as an exercise.

To find factors of your cubic,

$$f(x) = x^3-x^2-14x+24$$

We will try to find values of $x$ for which $f(x)=0$. Supposing that the polynomial can be neatly factored into integer factors $x-a$, $x-b$, and $x-c$, we know that $abc=24$, so it is beneficial to try positive and negative factors of $24$.

Let's try $1$, the smallest:

$$\begin{align} f(1)&=1^3-1^2-14\cdot1+24 \\ &= 10 \end{align}$$

Hmm, no luck. Next let's try $-1$:

$$\begin{align} f(1)&=(-1)^3-(-1)^2-14(-1)+24 \\ &= 36 \end{align}$$

Again we have not reached zero. Next let's try $2$:

$$\begin{align} f(2)&=2^3-2^2-14\cdot2+24 \\ &= 0 \end{align}$$

We've found a solution! This means that $x-2$ is a factor of the cubic. To find the remaining factors, we can do polynomial division to find some quadratic $x^2+bx+c$ so that $f(x) = (x-2)(x^2+bx+c)$.