Polynomial equation.

First we find the degree of $p$ , assume that the degree is $n$ so we can write : $p(x)= \sum_{k=0}^{n}a_k x^k\ \ \ \ \ \ a_n\neq 0,\ \ \forall k\in\{0,1\dots,n\}, a_k\in \mathbb{R}$ Then the leading coefficient in $(x+10)p(2x)$ is $2^n a_n$ . And the leading coefficient in $(8x-32)p(x+6)$ is $8 a_n$ , then : $8a_n=2^n a_n \overset{a_n\neq 0}{\Longleftrightarrow} 2^n =8 \Longleftrightarrow n=3.$ So $p$ at most three real roots : set $x=4$ to get $(4+10)p(8)=0\Longrightarrow p(8)=0$ set $x=-10$ to get $0=(-80-32)p(-4)\Longrightarrow p(-4)=0$ set $x=-2$ to get : $(-2+10)p(-4)=(-16-32)p(4) \Longrightarrow p(4)=0$ Then : $p(x)=c(x-4)(x+4)(x-8)$ , and $p(1)=210=105c$ then $c=2$ , therefore $p(x)=2(x-4)(x+4)(x-8)$ Easily : $p(10)= 2\cdot 6 \cdot 14 \cdot 2= \boxed{336}$