Number of real roots $x^8-x^5+x^2-x+1=0$ 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.

Another way: note that
$$
x^8-x^5+x^2-x+1=\left(x^4-\frac{1}{2}x\right)^2+\frac{1}{2}x^2+\left(\frac{1}{2}x-1\right)^2>0.
$$
Equality cannot occur since $x^4-\frac{1}{2}x$, $x$ and $\frac{1}{2}x-1$ cannot vanish together.
