How to show that a polynomial does not have real roots?

Clearly there is no negative root as all terms are positive for $x < 0$. The question remains if there are positive roots. Here is a simple way which often works.

Case 1: $0 < x <1$.

$$P(x) = (15-x) + (x^2-x^7) + x^8 > 0$$ as each term is positive.

Case 2: $ x > 1$. Similarly $$P(x) = (x^8-x^7) + (x^2-x) + 15 > 0$$

as $x=1$ is not a root, we are done.

The polynomial can be rewritten $$ x^7(x-1) + x(x-1) + 15. $$

Unless $x$ is between $0$ and $1$, the first two terms are positive, and so the polynomial is positive.

Even if $x$ is between $0$ and $1$, the first two terms are tiny in magnitude, certainly each individually greater than $-1$, so that when $15$ is added to their sum, the result is positive.

Thus the polynomial has no real roots.

It should be clear that on the interval $[-1,1]$ you have $|x^8-x^7+x^2-x|\leq |x^8|+|x^7|+|x^2|+|x|\leq 4$ and so $x^8-x^7+x^2-x+15\geq 11$

Further you should notice that $x^8-x^7>0$ when $|x|>1$ and that $x^2-x>0$ when $|x|>1$, so $x^8-x^7+x^2-x+15\geq 11$ for all $x$