Show that the polynomial $x^8 -x^7+x^2-x+15$ has no real root I am not getting on how to approach this problem. Clearly, this polynomial can have atleast 2 real roots. And using Descartes's rule of signs, it can have a maximum of 4 positive real roots. But after that, how should I proceed ?
Any help would be highly appreciated...

We have that $x^8-x^7$ and $x^2-x$ are bigger then zero for $x<0$ because $x^8>0$ and $-x^7>0$ similar logic for $x^2$ and $-x$ so $p(x)>15$ for $x<0$.
For $0<x<1$ we have that $x^8-x^7+x^2-x+15>-x^7-x+15>-1-1+15>13$ so $p(x)>13$
For $x\geq1$ we have that $x^8\geq x^7$ and $x^2\geq x$ so $x^8-x^7+x^2-x+15\geq 15$ so $p(x)\geq 15$.
Since $x=0$ is not a root we can conclude there aren't any roots.
