Show that the polynomial $x^{8}-x^{7}+x^{2}-x+15$ has no real root. 
Show that the polynomial $$x^{8}-x^{7}+x^{2}-x+15$$ has no real root.
Source

As I have learnt from my previous post,
using Descarte's Sign rule I am getting $4$ positive and $0$ negative roots.
So no of nonreal roots are $N-(p+q) = 8-(4+0)= 4$.So rest $4$ roots must be real! It is violating the question's condition.Can anybody help me out! I am not understanding the derivative concept regarding this!
 A: If $x<0$ every term is positive, hence the polynomial is $>0$.
If $x>1$ , then $x^2-x>0$ , $x^8 - x^7 >0$ , $15>0$ hence the function is $>0$.
If $0<x<1$, then $1-x>0$ , $x^2 - x^6>0$, $14+x^8>0$ hence the function is $>0$
Hence for every real $x$ the function is $>0$ , hence no real root.
A: Descarte's Sign rule is saying that you can have $4$,$2$ or $0$ positive real roots. The change of sign is giving the upper limit to the number of real roots not the exact value.
So it is not really helpful in deciding if there are no real roots at all. You would still have to argue that the polynomial has no negative evaluation at all.
So, it is very easy to see that this polynomial is always positive. For $|x|>1$ it is always $x^{2n}-x^{n} > 0$, because it is the same as $x^{2n} > x^{n}$, and this covers two parts of it. For $|x| \leq 1$, the sum $|x^{8}-x^{7}+x^{2}-x^{1}|$ cannot exceed $15$ as it is $4$ at worst.
The polynomial is then positive all the time.
A: $f(x)=x^8-x^7+x^2-x+15$ has atmost $4$ positive real roots due to $4$ sign changes in its coefficients.
$f(-x)=x^8+x^7+x^2+x+15$ has no sign changes so no negative real root.
(using Descartes rule of signs)
Further, $f(x)=(x-1)(x^7+x)+15$ and observe that $f(x)>0$ for all $x>0$ so there will be no positive root in $(0,\infty)$.
