# How do you find all zeros with multiplicities of f(x)=x^6-3x^3-10?

Dec 9, 2017

The six zeros are:

$x = \sqrt{5}$

$x = - \frac{1}{2} \sqrt{5} \pm \frac{1}{2} \sqrt{3} \sqrt{5} i$

$x = - \sqrt{2}$

$x = \frac{1}{2} \sqrt{2} \pm \frac{1}{2} \sqrt{3} \sqrt{2} i$

all with multiplicity $1$

#### Explanation:

${x}^{6} - 3 {x}^{3} - 10 = \left({x}^{3} - 5\right) \left({x}^{3} + 2\right)$

Then:

${x}^{3} - 5 = \left({x}^{3} - {\left(\sqrt{5}\right)}^{3}\right)$

$\textcolor{w h i t e}{{x}^{3} - 5} = \left(x - \sqrt{5}\right) \left({x}^{2} + \sqrt{5} x + \sqrt{25}\right)$

$\textcolor{w h i t e}{{x}^{3} - 5} = \left(x - \sqrt{5}\right) \left({\left(x + \frac{1}{2} \sqrt{5}\right)}^{2} + {\left(\frac{1}{2} \sqrt{3} \sqrt{5}\right)}^{2}\right)$

$\textcolor{w h i t e}{{x}^{3} - 5} = \left(x - \sqrt{5}\right) \left({\left(x + \frac{1}{2} \sqrt{5}\right)}^{2} - {\left(\frac{1}{2} \sqrt{3} \sqrt{5} i\right)}^{2}\right)$

$\textcolor{w h i t e}{{x}^{3} - 5} = \left(x - \sqrt{5}\right) \left(x + \frac{1}{2} \sqrt{5} - \frac{1}{2} \sqrt{3} \sqrt{5} i\right) \left(x + \frac{1}{2} \sqrt{5} + \frac{1}{2} \sqrt{3} \sqrt{5} i\right)$

and:

${x}^{3} + 2 = \left({x}^{3} + {\left(\sqrt{2}\right)}^{3}\right)$

$\textcolor{w h i t e}{{x}^{3} + 2} = \left(x + \sqrt{2}\right) \left({x}^{2} - \sqrt{2} x + \sqrt{4}\right)$

$\textcolor{w h i t e}{{x}^{3} + 2} = \left(x + \sqrt{2}\right) \left({\left(x - \frac{1}{2} \sqrt{2}\right)}^{2} + {\left(\frac{1}{2} \sqrt{3} \sqrt{2}\right)}^{2}\right)$

$\textcolor{w h i t e}{{x}^{3} + 2} = \left(x + \sqrt{2}\right) \left({\left(x - \frac{1}{2} \sqrt{2}\right)}^{2} - {\left(\frac{1}{2} \sqrt{3} \sqrt{2} i\right)}^{2}\right)$

$\textcolor{w h i t e}{{x}^{3} + 2} = \left(x + \sqrt{2}\right) \left(x - \frac{1}{2} \sqrt{2} - \frac{1}{2} \sqrt{3} \sqrt{2} i\right) \left(x - \frac{1}{2} \sqrt{2} + \frac{1}{2} \sqrt{3} \sqrt{2} i\right)$

So the six zeros are:

$x = \sqrt{5}$

$x = - \frac{1}{2} \sqrt{5} \pm \frac{1}{2} \sqrt{3} \sqrt{5} i$

$x = - \sqrt{2}$

$x = \frac{1}{2} \sqrt{2} \pm \frac{1}{2} \sqrt{3} \sqrt{2} i$