TITLE: Prove that a polynomial has at least one nonreal complex root
QUESTION [15 upvotes]: Prove that the polynomial below has at least one nonreal complex root
$$x^5+\frac{x^4}2+ \frac{x^3}3+\frac{x^2}4+\frac x{24}+\frac 1{120}$$
I have tried to prove that there exist $k\in \Bbb R$, such that the above polynomial can be factored into $(x^2+k)P(x)$, where deg$(P)=3$. But somehow I couldn't work it out.

REPLY [15 votes]: Here is a way to do it without resorting to calculus:
Eliminate the quartic term (making it into a "depressed quintic") by making the substitution $x=z-1/10$ (as $1/10$ is one fifth of the coefficient of $x^4$), which turns the polynomial into $$z^5+\frac{7 z^3}{30}+\frac{17 z^2}{100}+\frac{z}{6000}+\frac{239}{37500}\text{.}$$ By Descartes' Rule of Signs, this has no positive roots and either three or one negative roots. $0$ certainly isn't a root, so it must have at least 2 non-real roots. 

As an aside, you could use the discriminant to show further that it must have exactly one real root, but the discriminant is $2258539/17915904000$ and is not something you could calculate by hand, given the formula for the discriminant of a monic quintic is pretty terrible (which you can verify with Wolfram|Alpha).