# How do you find all the zeros of f(x) = x^3 - 6x^2 - x + 6 ?

Mar 23, 2016

Factor by grouping to find zeros: $\left\{\begin{matrix}x = 1 \\ x = - 1 \\ x = 6\end{matrix}\right.$

#### Explanation:

Factor by grouping:

${x}^{3} - 6 {x}^{2} - x + 6$

$= \left({x}^{3} - 6 {x}^{2}\right) - \left(x - 6\right)$

$= {x}^{2} \left(x - 6\right) - 1 \left(x - 6\right)$

$= \left({x}^{2} - 1\right) \left(x - 6\right)$

$= \left(x - 1\right) \left(x + 1\right) \left(x - 6\right)$

So the zeros are:

$\left\{\begin{matrix}x = 1 \\ x = - 1 \\ x = 6\end{matrix}\right.$