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.

If $x\geq1$ so
$$x^8-x^5+x^2-x+1=(x^8-x^5)+(x^2-x)+1>0.$$
If $0\le x<1$ so
$$x^8-x^5+x^2-x+1=1-x+x^2(1-x^3)+x^8>0.$$
If $x<0$ so after replacing $x$ on $-x$ we obtain:
$$x^8+x^5+x^2+x+1$$ has no positive roots by the Descarte's rule.
