Determine all polynomials $P$ satisfying the following equation for all real $x$:

$$
\left(x^{2}-6 x+8\right) P(x)=\left(x^{2}+2 x\right) P(x-2)
$$
First, note that the zero polynomial is a solution to the equation in the statement.

Now suppose that $P$ is a solution to the equation and that $P$ is non-zero. Let's first rewrite the equation from the hypothesis. It is indeed equivalent to

$$
(x-2)(x-4) P(x)=x(x+2) P(x-2)
$$

By substituting $x$ with 0, we get $8 P(0)=0$ which means $P(0)=0$. By substituting $x$ with -2, we get $24 P(-2)=0$ which means $P(-2)=0$. By substituting $x$ with 4, we have $0=24 P(2)$ which means $P(2)=0$. The polynomial $P$ thus has at least 3 roots: $-2, 0, 2$. Therefore, $x, x-2$, and $x+2$ all divide the polynomial $P$, so there exists a polynomial $Q$ such that $P(x)=x(x-2)(x+2) Q(x)$ for all real $x$. By replacing $P$ with $x(x-2)(x+2) Q(x)$ in the equation, we get

$$
(x-2)(x-4) x(x+2)(x-2) Q(x)=x(x+2)(x-4) x(x-2) Q(x-2)
$$

which is equivalent to

$$
(x-4)(x+2)(x-2) x[(x-2) Q(x)-x Q(x-2)]=0
$$

Thus, for all $x$ different from $-2, 0, 2, 4$, we have $(x-2) Q(x)-x Q(x-2)=0$. The polynomial $R(x)=(x-2) Q(x)-x Q(x-2)$ has an infinite number of roots (it suffices to take all integers different from $-2, 0, 2, 4$), so it is the zero polynomial. Therefore, for all real $x$, we have $(x-2) Q(x)-x Q(x-2)=0$. In particular, for $x=0$, we have $-2 Q(0)=0$ so $Q(0)=0$ and 0 is a root of the polynomial $Q$ so $x$ divides $Q$. Therefore, there exists a polynomial $S$ such that $Q(x)=x S(x)$ and the equation becomes

$$
(x-2) x S(x)=x(x-2) S(x-2)
$$

or equivalently

$$
x(x-2)[S(x-2)-S(x)]=0
$$

so for $x$ different from 0 and 2, we have $S(x-2)=S(x)$. In particular, by immediate induction, we find that $S(x)=S(4)$ for all even integers strictly greater than 2. By setting $c=S(4)$, we see that the polynomial $T$ defined by $T(x)=S(x)-c$ has an infinite number of roots, so it is the zero polynomial. Therefore, $S(x)=c$ for all real $x$, which implies that $P(x)=x(x-2)(x+2) x S(x)=x^{2}\left(x^{2}-4\right) c$ with $c$ a non-zero real number. Conversely, we verify that this polynomial is indeed a solution to the polynomial equation since we have

$$
(x-2)(x-4) P(x)=(x-2)(x-4) x^{2}\left(x^{2}-4\right) c=x(x+2) P(x-2)
$$

The solutions are therefore all polynomials $P$ defined by $P(x)=x^{2}\left(x^{2}-4\right) c$ with $c$ a real number, and these are the only solutions, noting that the zero polynomial corresponds to the case where $c=0$.