Dear Uncle Colin,

I need to factorise $x^3 - 4x^2 - 60x + 288$. What’s your best approach?

- Cubics Are Really Dangerous And Nasty

Hi, CARDAN, and thanks for your message!

That’s an awkward one: 288 has a lot of factors!

My best approach is to differentiate first and find turning points ((In this case we get lucky and one of the turning points turns out to be a double root; in general, the derivative approach helps to get an idea of the shape of the graph and give some bounds on roots.)): the derivative is $3x^2 - 8x - 60$, which factorises as $(3x-10)(x-6)$ – so there are turning points at $x=\frac{10}{3}$ and $x=6$.

Working from smallest to largest, when $x = \frac{10}{3}$, the expression gives something horrible, $\frac{1000}{27} - \frac{400}{9} - \frac{600}{3} + 288$; that’s $-\frac{200}{27} + 88$, which is positive. This means that there’s exactly one ((When x is a large negative number, the value is large and negative; this is the leftmost turning point, and the function is continuous, so it can only cross the axis once.)) root where $x < \frac{10}{3}$.

When $x = 6$, the expression gives $216 - 144 - 360 + 288$, which is zero. That means there’s a double root at $x=6$ and $(x-6)^2$ is a factor.

The other factor must be $(x+8)$ because $288 = 36 \times 8$. It’s worth checking that: $-512 - 256 + 480 + 288$ does indeed give zero.

Your expression is $(x+8)(x-6)^2$. Hope that helps!

- Uncle Colin