# How do you find all zeros of the function f(x)= x^4 + 5x^2 + 4 given -i as a zero?

Apr 6, 2016

Factor by grouping to find:

$x = \pm i$ or $x = \pm 2 i$

#### Explanation:

This polynomial factors by grouping as follows:

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

$= {x}^{4} + 4 {x}^{2} + {x}^{2} + 4$

$= {x}^{2} \left({x}^{2} + 4\right) + 1 \left({x}^{2} + 4\right)$

$= \left({x}^{2} + 1\right) \left({x}^{2} + 4\right)$

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

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

Hence zeros: $x = \pm i$ and $x = \pm 2 i$

$\textcolor{w h i t e}{}$
Another way of finding this is to note that since $f \left(x\right)$ has Real coefficients, any zeros must occur in Complex conjugate pairs.

So if $x = - i$ is a zero then so is $x = i$

Therefore $f \left(x\right)$ is divisible by $\left(x + i\right) \left(x - i\right) = {x}^{2} - {i}^{2} = {x}^{2} + 1$

Then dividing $f \left(x\right)$ by ${x}^{2} + 1$ we find:

$f \left(x\right) = \left({x}^{2} + 1\right) \left({x}^{2} + 4\right)$

Hence the other two zeros are $\pm 2 i$