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 $\qquad$
$$
f(x)=(x+1)(x-1)(x-3)(x-5)=\left(x^{2}-4 x-5\right)\left(x^{2}-4 x+3\right)=\left[\left(x^{2}-4 x\right)-1\right]^{2}-16 \text {, }
$$

So when $x^{2}-4 x-1=0$, the minimum value of $f(x)$ is -16.