# How do you find all zeros with multiplicities of f(x)=x^5-2x^4-4x+8?

May 29, 2017

The zeros are $2$, $\pm \sqrt{2}$ and $\pm \sqrt{2} i$

#### Explanation:

Given:

$f \left(x\right) = {x}^{5} - 2 {x}^{4} - 4 x + 8$

Note that the ratio between the first and second terms is the same as that between the third and fourth terms, so we can factor this quadrinomial by grouping:

${x}^{5} - 2 {x}^{4} - 4 x + 8$

$= \left({x}^{5} - 2 {x}^{4}\right) - \left(4 x - 8\right)$

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

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

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

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

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

$= \left(x - \sqrt{2}\right) \left(x + \sqrt{2}\right) \left(x - \sqrt{2} i\right) \left(x + \sqrt{2} i\right) \left(x - 2\right)$

Hence zeros:

$\pm \sqrt{2}$, $\pm \sqrt{2} i$, $2$