9. Given that $F(x)$ represents a fifth-degree polynomial in $x$. If
$$
\begin{array}{l}
F(-2)=F(-1)=F(0)=F(1)=0, \\
F(2)=24, F(3)=360,
\end{array}
$$

then the value of $F(4)$ is (.
(A) 1800
(B) 2011
(C) 4020
(D) cannot be determined
9. A.

Let $F(x)=x(x+2)(x+1)(x-1)(a x+b)$.
Then $\left\{\begin{array}{l}24(2 a+b)=24, \\ 120(3 a+b)=360 .\end{array}\right.$
Solving, we get $a=2, b=-3$.
Thus, $F(4)=1800$.