# How do I find all the rational zeros of p(x)=x^3-12x-16?

Check if any of the divisors of $- 16$ is a root of $p \left(x\right)$ in our case

$x = 4$ is a root hence $p \left(4\right) = 0$

The polynomial can be written as

$p \left(x\right) = \left(x - 4\right) \cdot \left({x}^{2} + b x + c\right)$

It is easy to proove that $b = 4$ and $c = 4$

Hence

$p \left(x\right) = \left(x - 4\right) {\left(x + 2\right)}^{2}$

So the zeros are $x = 4$ ,$x = - 2$ (double root)