TITLE: Number of real roots $x^8-x^5+x^2-x+1=0$
QUESTION [5 upvotes]: Find the number of real roots of $x^8-x^5+x^2-x+1$.
My attempt: $f(x)$ has $4$ sign changes and $f(-x)$ has no sign changes, so the possibility of having real roots is $4+0=4$. Since this is a polynomial of degree $8$ it should have $8-4=4$ imaginary roots.
It is quite impossible to see that this equation does not have any real roots by observing the factor $x^8-x^5+x^2-x+1= (x-1)(x^7+x^6+x^5+x)+1>0$, hence it can't have any real roots. My question is how to prove using this equation doesn't have any real roots using Descartes' Rule? Please give some useful hints.

REPLY [2 votes]: Just another way, using AM-GM:
$$(x^8+\tfrac12x^2) + (\tfrac12x^2+1)> \sqrt2|x^5|+\sqrt2|x| \geqslant x^5+x$$