# How do you find all zeros with multiplicities of f(x)=3x^4-14x^2-5?

Feb 13, 2017

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

$x = \pm \sqrt{5}$

#### Explanation:

$f \left(x\right) = 3 {x}^{4} - 14 {x}^{2} - 5$

$\textcolor{w h i t e}{f \left(x\right)} = \left(3 {x}^{4} - 15 {x}^{2}\right) + \left({x}^{2} - 5\right)$

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

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

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

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

Hence zeros:

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

$x = \pm \sqrt{5}$