First substitute $x=2,-2$ then we can get the fact that: $P(2)=P(0)=P(-2)$ so let $P(x)=(x-2)x(x+2)f(x)+C$ then $(x-2)x(x+2)(x+4)f(x+2)+(x+2)x(x-2)(x-4)f(x-2)=2(x-2)x^2(x+2)f(x)$ so $(x+4)f(x+2)+(x-4)f(x-2)=2xf(x)$ then substitute $x-4$ instead of $x$ then we get: $(x-4)f(x-2)-xf(x)=xf(x)-(x+4)f(x+2)$ by substituting $x+4k$ instead of $x$ where $k$ is integer we get infinite number of such cases then by letting $(x-4)f(x-2)-xf(x)=g(x)$ we should have infinite case of solutions for specific value then we conclude that $f(x)$ is constant function so $P(x)=c_1(x-2)x(x+2)+c_2$ for arbitrary real number $c_1, c_2$

Skravin wrote: $(x-4)f(x-2)-xf(x)=xf(x)-(x+4)f(x+2)$ by substituting $x+4k$ instead of $x$ where $k$ is integer we get infinite number of such cases then by letting $(x-4)f(x-2)-xf(x)=g(x)$ we should have infinite case of solutions for specific value then we conclude that $f(x)$ is constant function Why? From this equation we can only get $g(x)\ne g(x+2)$,$g(x)\ne g(x+4)$

smy2012 wrote: Skravin wrote: $(x-4)f(x-2)-xf(x)=xf(x)-(x+4)f(x+2)$ by substituting $x+4k$ instead of $x$ where $k$ is integer we get infinite number of such cases then by letting $(x-4)f(x-2)-xf(x)=g(x)$ we should have infinite case of solutions for specific value then we conclude that $f(x)$ is constant function Why? From this equation we can only get $g(x)\ne g(x+2)$,$g(x)\ne g(x+4)$ Well I think I took the wrong form to express my idea... just use $(x+4)f(x+2)+(x-4)f(x-2)=2xf(x)$ directly by substituting $x=0$ we get $f(2)=f(-2)$ by substituting $x=4$ we get $f(6)=f(4)$ by substituting $x=6$ we get $f(8)=f(6)$ then we get the conclusion that substituting $x+2$ instead of $x$ continuously where $f(x)=f(x+2)$ for specific $x$ we can always get other form of $f(x)=f(x+2)$(because if $f(x)=f(x+2)$ then $f(x+6)f(x+4)=(2x+4-(x-2))f(x+2)$

Let $Q\in \mathbb{R}[x]$ that $Q(x)= P(x+2)-P(x)$. We have $(x-2)Q(x)\equiv (x+2)Q(x-2)$. Easy to see that $Q(0)=Q(-2)=0$. Let $R\in \mathbb{R}[x]$ be a polynomial that $Q(x)= x(x+2)R(x)$. So $R(x)\equiv R(x+2)\implies R\equiv c$ for a real constant $c$. If $c=0$, we get $Q(x)= 0$ and $P(x+2)= P(x)$. So $P$ is constant polynomial. If $c\neq 0$, we get $cx(x+2)= P(x+2)-P(x)$. Let $\deg (P)=n\in \mathbb{Z}^+$ and $P(x)= a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0$. The coefficient of $x^{n-1}$ in $P(x+2)-P(x)$ is $2na_n\neq 0$. So, $\deg (P(x+2)-P(x))= n-1$. Thus $2=n-1\implies n=3$. We have $cx(x+2)=6a_3x^2+(12a_3+4a_2)x+(8a_3+4a_2+2a_1)\implies a_2=0,a_1=-4a_3$. Finally, the answer is $P(x)=a_3(x^3-4x)+a_0$ where $a_3$ and $a_0$ are arbitrary real numbers.

We can find easily that $P(0)=P(2)=P(-2)$. Suppose they are all equal to $c$. Then, $P(x)-c$ satisfies the given condition also. Without loss of generalization we assume $P(0)=P(2)=P(-2)=0$. If we keep plugging $2n$ into $x$, we can see recursively that $P(2n)=\frac{n\left(n^{2}-1\right)}{6}P(4)$. So $P(x)=dx\left(x^{2}-4\right)$, and we can find easily that this satisfies the given condition. Therefore, the full answer is \[P(x)=a\left(x^{3}-4x\right)+b\]