Let $ p(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_0, a_n\not =0$, then $ a_n2^nx^{n+1}+...=8a_nx^{n+1}\to n=3$. $ (2x-8)|p(2x)\to (x-8)|p(x)$, $ (x+10)|p(x+6)\to (x+4)|p(x)$. Therefore $ p(x)=(x+4)(x-8)(ax+b)$ and $ (x+10)(2x+4)(2x-8)(2ax+b)=(8x-32)(x+10)(x-2)(ax+6a+b)$. It give $ a(x+2)(x+b/2a)=a(x-2)(x+6+b/a)$ or $ b=-4a$ or $ p(x)=a(x-4)(x-8)(x+4)=a(x-8)(x^2-16)$. From $ p(1)=a(-15)(-7)=210\to a=2$ or $ p(x)=2(x-8)(x^2-16).$

I have reached the same result ..but I don't understand how did you prove that the polynomial was of degree 3...I used the fact that if I had another root I would have infinit roots reaching a contradiction...your solutions seems to be simpler and I would like you to explain me that .Thank you!! PD: First post!

Look at the leading coefficient. On the left it is 2^n, and on the right it is 8, so n=3

Sorry for bringing up an old post but, I have made an easy to read pdf file of the solution(In spanish) and I just want to post it, it doesn't contain any accents.