[Solution] Substituting $x=0,2$ into the given identity, we can see that the polynomial $P(x)$ has roots 0 and 1, meaning it is divisible by the polynomial $x^{2}-x$. Next, substituting $P(x) \equiv\left(x^{2}-x\right) Q(x)$ into the identity, we get
$$
\begin{array}{l}
x\left[(x-1)^{2}-(x-1)\right] Q(x-1) \equiv(x-2)\left(x^{2}-x\right) Q(x), \\
x(x-1)(x-2) Q(x-1) \equiv x(x-1)(x-2) Q(x) .
\end{array}
$$

Thus, the polynomial $Q(x)$ satisfies the identity
$$
Q(x) \equiv Q(x-1) .
$$

This implies $Q(0)=Q(1)=Q(2)=\cdots$
Therefore, $Q(x) \equiv a$, where $a$ is a constant, and the desired polynomial is
$P(x)=a\left(x^{2}-x\right)$. Conversely, it is easy to verify that all such polynomials also satisfy the given identity.
5. 138 Find all polynomials \( P(x) \) that satisfy
\[
x P(x-1) \equiv (x-2) P(x), \quad x \in \mathbb{R}
\]