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.

We will prove that $f(x)=x^8-x^5+x^2-x+1$ has no real roots using Descartes' Rule only.
Since $f(-x)=x^8+x^5+x^2+x+1$ has no non-negative coefficients we know $f(x)$ has no real roots in $(-\infty,0]$.
Since $f(x+1)=x^8+8x^7+28x^6+55x^5+65x^4+46x^3+19x^2+4x+1$ has no non-negative coefficients we know $f(x+1)$ has no real roots in $[0,\infty)$. This means $f(x)$ has no real roots in $[1,\infty)$.
It suffices to show that $f(x)$ has no real roots in $(0,1)$. Perform the substitution $t=1/x$ so it suffices to show that $g(t)=t^8f(t)=t^8-t^7+t^6-t^3+1$ has no real roots in $(1,\infty)$.
Since $g(t+1)=t^8+7t^7+22t^6+41t^5+50t^4+40t^3+19t^2+4t+1$ has no non-negative coefficients we know $g(t+1)$ has no real roots in $(0,\infty)$. This means that $g(t)$ has no real roots in $(1,\infty)$, so $f(x)$ has no real roots in $(0,1)$. This completes the proof.
