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.
 A: 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.
A: 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.
A: Just another way, using AM-GM:
$$(x^8+\tfrac12x^2) + (\tfrac12x^2+1)> \sqrt2|x^5|+\sqrt2|x| \geqslant x^5+x$$
A: $f(x)=x^8-x^5+x^2-x+1=0$ has no real root as $f(x)>0$ for all real values of $x$.
$$f(x)=x^3(x^3-1)+x(x-1)+1 >0, ~if~ x>1$$
$$f(x)=x^8+x^2(1-x^3)+(1-x) > ~if~x <1$$
And $f(1)=1>0$.
A: 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.
