Find polynomials $p(x) $ such that $(x+10)p(2x)=(8x-32)p(x+6) $ when $p(1)=210$ Hi I have polynomials as below form.
$(x+10)p(2x)=(8x-32)p(x+6)$, when $p(1)=210$
I try to assign some $x$ eg. $0.5$ , $-6$ but can not find the correct relation. Any advice or guidance on this would be greatly appreciated, Thanks.

Let $S$ be the set of zeroes of $p$.
If $x \in S$, $x \ne -4$ then $$\underbrace{(x+4)}_{\ne 0}p(2x-12) = 8(x-10)p((x-6)+6) = 8(x-10)p(x) = 0$$
so $2x-12 \in S$.
Notice that $8 \in S$ since plugging in $x = 4$ gives $14p(8) = 8(4-4)p(10) = 0$.
Therefore $4 = 2\cdot 8 -12 \in S$ and then $-4 = 2\cdot 4 - 12 \in S$ so $\{8,-4,4\} \subseteq S$.
@Vasily Mitch noticed that $\deg p = 3$ so the only solution is $S = \{8,-4,4\}$ or $$p(x) = a(x-8)(x-4)(x+4)$$
for some $a \in \mathbb{R}$. From $p(1) = 210$ it follows $a = 2$.
