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

Question

1561 views around the world

You can reuse this answer
Creative Commons License

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.