Determine all real-coefficient polynomials $p(x)$ such that
$$
t s^{\prime}(t \cdots 1)=(t-2) p(t)
$$

holds for all real numbers $t$.
For $t=2, 2 p(1)=0$, i.e., 1 is a $\cdots$ root of $p(x)$. For $t=1, p(0)=-p(1)=0$. Thus, 0 is a root of $p(x)$. Therefore, we can write $p(x)$ as $x(x-1) q(x)$, where $q(x)$ is a real polynomial. Substituting $p(t)=t(t-1) q(t)$ into the given equation,

\[
\begin{array}{l}
t(t-1)(t-2) q(t-1) \\
=(t-2) t(t-1) q(t), \\
q(t-1)=q(t), \quad t \neq 0,1,2.
\end{array}
\]

Let $s(x)=q(x)-q(3)$, we can deduce that $s(n)=0$, where $n=3,4,5, \cdots$. The polynomial $s(x)$ has infinitely many roots, so $s(x)$ is the zero polynomial, $q(x)=q(3)$ is a constant polynomial. Therefore, the required polynomial must be of the form $c x(x-1)$, where $c \in \mathbb{R}$.

On the other hand, if $c \in \mathbb{R}, p(x)=c x(x-1)$ clearly satisfies the given equation. Therefore, the answer to the problem is $p(x)=c x(x-1)$, $c \in \mathbb{R}$.