# How do you find the number of complex zeros for the function f(x)=5x^5+36x^3+7x?

Oct 5, 2016

There are four complex zeros

#### Explanation:

First of all, we can factor out an $x$ factor, so there's surely a root for $x = 0$:

$5 {x}^{5} + 36 {x}^{3} + 7 x = x \left(5 {x}^{4} + 36 {x}^{2} + 7\right)$

For the remaining part, let $t = {x}^{2}$. This leads us to

$5 {x}^{4} + 36 {x}^{2} + 7 = 5 {t}^{2} + 36 t + 7$

You can verify that this quadratic equation has solutions $t = - \frac{1}{5}$ and $t = - 7$.

Substituting back, we'd get

$t = - \frac{1}{5} \setminus \implies {x}^{2} = - \frac{1}{5} \setminus \implies x = \pm \frac{i}{\sqrt{5}}$

$t = - 7 \setminus \implies {x}^{2} = - 7 \setminus \implies x = \pm i \sqrt{7}$

This means that there are four complex roots.