Ubiquitous +

$\begin{aligned} f(x) & = x^8 - x^7 + x^2 - x + 15 \\ & = \left(x^6+1\right)\left(x^2-x \right) + 15 \end{aligned}$

We note that $f(x)$ is minimum, when $\left(x^6+1\right)\left(x^2-x \right)$ is minimum. And $\left(x^6+1\right)\left(x^2-x \right)$ is minimum, when $\left(x^2-x \right) < 0$ or $0<x<1$ .

$\begin{aligned} \left(\color{#3D99F6}{x^6}+1\right)\left(x^2-x \right) & > \left(\color{#3D99F6}{1}+1\right)\left(x^2-x \right) \quad \quad \small \color{#3D99F6}{\text{Since }x < 1} \\ & > -2\color{#3D99F6}{x(1-x)} \quad \quad \small \color{#3D99F6}{\text{By AM-GM inequality: }x(1-x) = \left(\frac {x+1-x}2 \right)^2 = \frac 14} \\ & > - \frac 12 \quad \quad \small \color{#3D99F6}{\text{Equality occurs when }x = \frac 12.} \end{aligned}$

$\implies f(x) > 14.5$ and therefore, $f(x)$ has $\boxed{0}$ real root.

$f(x) = x^7(x-1)+x(x-1)+15 = x(x-1)(x^6+1)+15$

Now, $\mathbb{R} = (- \infty , 0] \cup (0,1] \cup (1, \infty)$

Note that, $f(0)=f(1)=15$

When $x \in (1, \infty),$ $x,x-1 \ and \ x^6+1 > 0 \implies f(x) > 15 \ \forall \ x \in (1, \infty)$

When $x \in (- \infty , 0),$ $x \ and \ x-1 < 0 \ thus \ x(x-1) > 0$ since $x^6+1 > 0$ for any $x,f(x) > 15$ in this case too

When $x \in (0,1),$ $1 < x^6+1 < 2$ and $0 < x < 1$ . Since both of them are positive $0 < x(x^6+1) < 2$ . Further, $-1 < x-1 < 0,$ thus $x(x-1)(x^6+1) < 0$ . Again $|x-1| < 1$ this implies $-2 < x(x-1)(x^6+1) < 0$ . Thus $f(x) > 13 > 0$ .

Combining all these cases we see that $f(x) > 0 \ \forall \ x \in \mathbb{R}$ which shows $f(x)$ has no real roots.