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 $p(x)=x^8-x^7+x^2-x+15=x(x-1)(x^6+1)+15$
If $x \le 0$ then $x-1<0$, hence $x(x-1) \ge 0$, which gives $p(x)>0$.
If $x \ge 1$, then it is clear that $p(x)>0$.
Now let $0<x<1$. Then $x(x-1) \ge -1/4$, hence $x(x-1)(x^6+1) \ge -\frac{1}{4}(x^6+1)$.
Now it is your turn to show that  $-\frac{1}{4}(x^6+1)+15>0$ for $0<x<1$.
