2. If the function
$$
f(x)=\left(x^{2}-1\right)\left(x^{2}+a x+b\right)
$$

satisfies $f(x)=f(4-x)$ for any $x \in \mathbf{R}$, then the minimum value of $f(x)$ is $\qquad$ .
2. -16 .

Notice that, $f(1)=f(-1)=0$.
Also, $f(x)=f(4-x)$, so $f(3)=f(5)=0$.
Therefore, $f(x)=(x^{2}-1)(x-3)(x-5)$ $=(x^{2}-4x+3)(x^{2}-4x-5)$.
Let $t=x^{2}-4x+4 \geqslant 0$, then $f(x)=(t-1)(t-9)=(t-5)^{2}-16$.
Thus, the minimum value of $f(x)$ is -16.