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

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Answer 1 · 2016-10-05T19:46:26

There are four complex zeros

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

$5x^5+36x^3+7x = x(5x^4+36x^2+7)$

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

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

You can verify that this quadratic equation has solutions $t=-1/5$ and $t=-7$ .

Substituting back, we'd get

$t=-1/5 \implies x^2=-1/5 \implies x=pm i/sqrt(5)$

$t=-7\implies x^2=-7\implies x=pm isqrt(7)$

This means that there are four complex roots.

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