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...

Note that $x^8-x^7$ and $x^2 - x$ look very similar. We can use this to factor some of the terms:
$$
x^8 - x^7 + x^2 - x + 15\\
= x^7(x-1) + x(x-1) + 15\\
= (x^7+x)(x-1) + 15\\
= x(x^6+1)(x-1) + 15
$$
If the entire expression is to be zero, then at the very least, we must have $x(x^6+1)(x-1)< 0$, which happens for $x\in (0, 1)$. But for $x\in (0, 1)$, we have
$$
|x|<1\\
|x^6+1|<2\\
|x-1|<1
$$
which implies $|x(x^6-1)(x-1)|<2$, which in turn implies that $x(x^6+1)(x-1) + 15$ is always larger than $13$.
