# How do you find all the zeros of f(x) = x^3 - 8x^2 - x + 8?

Jul 2, 2016

Factor by grouping to find zeros:

$x = 1$, $x = - 1$, $x = 8$

#### Explanation:

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

Notice that the ratio of the first and second terms is the same as that between the third and fourth. So this cubic factors by grouping:

${x}^{3} - 8 {x}^{2} - x + 8$

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

$= {x}^{2} \left(x - 8\right) - 1 \left(x - 8\right)$

$= \left({x}^{2} - 1\right) \left(x - 8\right)$

$= \left(x - 1\right) \left(x + 1\right) \left(x - 8\right)$

Hence zeros: $x = 1$, $x = - 1$, $x = 8$