jmc

Setting $x=4,$ we get $$14p(8)=0,$$so $p(x)$ has a factor of $x-8.$

Setting $x=-10,$ we get $$8(-14)p(-4)=0,$$so $p(x)$ has a factor of $x+4.$

Setting $x=-2,$ we get $$8p(-4)=8(-6)p(4).$$Since $p(-4)=0,$ $p(4)=0,$ which means $p(x)$ has a factor of $x-4.$

Let $$p(x)=(x-8)(x-4)(x+4)q(x).$$Then $$(x+10)(2x-8)(2x-4)(2x+4)q(2x)=8(x-4)(x-2)(x+2)(x+10)q(x+6).$$This simplifies to $q(2x)=q(x+6).$

Let $q(x)=q_n{x}^n+q_{n-1}{x}^{n-1}+\cdots +q_{1}x+q_0.$ Then the leading coefficient in $q(2x)$ is $q_n{2}^n,$ and the leading coefficient in $q(x+6)$ is $q_n.$ Since $q(2x)=q(x+6),$ $$q_n{2}^n=q_n.$$Since $q_n\ne 0,$ $2^n=1,$ so $n=0.$ This means $q(x)$ is a constant polynomial. Let $q(x)=c,$ so $$p(x)=c(x-8)(x-4)(x+4).$$Setting $x=1,$ we get $$c(1-8)(1-4)(1+4)=210,$$so $c=2.$ Therefore, $p(x)=2(x-8)(x-4)(x+4),$ so $p(10)=2(10-8)(10-4)(10+4)=\boxed{336}.$