<s>[INST] 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.
 [/INST] 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.
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