Let's determine the polynomial $p(x)$, given that $p(12)=12$!, and for all $x$ it holds that

$$
x \cdot p(x-1)=(x-12) \cdot p(x)
$$
By substituting $x=0$ in the condition,

$$
0=(-12) \cdot p(0)
$$

thus, the polynomial has a root at zero. From the condition, it is also clear that if $(\alpha-1)$ is a root of the sought polynomial — then the left side is zero — and if $\alpha \neq 12$, then $p(\alpha)=0$ on the right side, so $\alpha$ is also a root. Together with 0, the numbers $1,2, \ldots, 11$ are also roots, so after factoring out the corresponding root factors,

$$
p(x)=x(x-1) \cdot(x-2) \cdot \ldots \cdot(x-11) \cdot q(x)
$$

Substituting the found form into the condition and simplifying by $x(x-1) \ldots(x-12)$, we get that for example for all $x$ greater than 12,

$$
q(x)=q(x-1)
$$

therefore for all such $x$,

$$
q(x-1)=q(x)=q(x+1)=q(x+2)+\ldots
$$

the polynomial $q$ thus takes, for example, the value $q(12)$ infinitely many times, and is therefore constant. Denoting its value by $c$,

$$
p(x)=c \cdot x(x-1) \cdot(x-2) \cdot \ldots \cdot(x-11)
$$

from which

$$
p(12)=c \cdot 12!, \quad \text { or } \quad c=1
$$

With this, we have shown that if the polynomial $p(x)$ satisfies both given conditions, then only

$$
p(x)=x(x-1) \cdot \ldots \cdot(x-11)
$$

is possible.

It is easily seen that for this polynomial, the conditions of the problem are indeed satisfied.