# How do you find the zeros of x^3-x^2-10x-8?

May 25, 2016

$x =$ Either -1 or -2 or +4

#### Explanation:

$f \left(x\right) = {x}^{3} - {x}^{2} - 10 x - 8$

To find the zeros of this polynomial function we must find all values of x such that $f \left(x\right) = 0$

Notice that $f \left(x\right)$ factorises to $\left(x + 1\right) \left({x}^{2} - 2 x - 8\right)$
Which further factorises to $\left(x + 1\right) \left(x + 2\right) \left(x - 4\right)$

Hence the zeors of $f \left(x\right)$ occur when $\left(x + 1\right) \left(x + 2\right) \left(x - 4\right) = 0$

This occurs when either $\left(x + 1\right) = 0$ or $\left(x + 2\right) = 0$ or $\left(x - 4\right) = 0$

Therefore the zeros of $f \left(x\right)$ are either -1 or -2 or +4