# How do you find the zeros of f(x) = x^4 - x^3 - 6x^2 + 4x + 8?

Jun 10, 2016

$x = - 1$

$x = 2$ with multiplicity $2$

$x = - 2$

#### Explanation:

First note that $f \left(- 1\right) = 1 + 1 - 6 - 4 + 8 = 0$.

So $x = - 1$ is a zero and $\left(x + 1\right)$ a factor:

${x}^{4} - {x}^{3} - 6 {x}^{2} + 4 x + 8 = \left(x + 1\right) \left({x}^{3} - 2 {x}^{2} - 4 x + 8\right)$

Note that in the remaining cubic, the ratio of the first and second terms is the same as that between the third and fourth terms. So this will factor by grouping:

${x}^{3} - 2 {x}^{2} - 4 x + 8$

$= \left({x}^{3} - 2 {x}^{2}\right) - \left(4 x - 8\right)$

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

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

$= \left(x - 2\right) \left(x + 2\right) \left(x - 2\right)$

Hence the other zeros are:

$x = 2$ with multiplicity $2$

$x = - 2$