# How do you find all zeros of f(x)=2x^4-2x^2-40?

Feb 5, 2017

$x = \pm \sqrt{5} \text{ }$ or $\text{ } x = \pm 2 i$

#### Explanation:

Given:

$f \left(x\right) = 2 {x}^{4} - 2 {x}^{2} - 40$

We can first treat this as a quadratic in ${x}^{2}$ then factor the two resulting quadratic factors using the difference of squares identity:

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

as follows:

$f \left(x\right) = 2 {x}^{4} - 2 {x}^{2} - 40$

$\textcolor{w h i t e}{f \left(x\right)} = 2 \left({\left({x}^{2}\right)}^{2} - {x}^{2} - 20\right)$

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

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

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

Hence zeros:

$x = \pm \sqrt{5} \text{ }$ or $\text{ } x = \pm 2 i$