6. 172 Determine all real-coefficient polynomials $p(x)$ such that
$$
t p(t-1)=(t-2) p(t)
$$

for all real numbers $t$.

保留了源文本的换行和格式。
[Solution] Let $t=2$, we get $2 p(1)=0$, so $p(1)=0$. Thus, $1$ is a root of the polynomial $p(x)$. Let $t=1$, we get $p(0)=-p(1)=0$, so $0$ is also a root of the polynomial $p(x)$. Let $p(x)=x(x-1) q(x)$, then
$$
\begin{array}{l}
t(t-1)(t-2) q(t-1)=(t-2) \cdot t(t-1) q(t) \\
q(t-1)=q(t), t \neq 0,1,2 .
\end{array}
$$

By setting $t=4,5,6, \cdots \cdots$, we get
$$
q(x)=q(3), x=4,5,6, \cdots
$$

That is, $q(x)-q(3)=0$ has infinitely many roots,
so $q(x) \equiv q(3)$.
Let $q(3)=c$, then we have
$$
p(x)=c x(x-1), c \in \mathbb{R}.
$$

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