2. Given $x_{1}=1, x_{2}=2, x_{3}=3$ are the three zeros of $f(x)=x^{4}+a x^{3}+b x^{2}+c x+d$, then $f(0)+f(4)=$
$\qquad$ .
Let the 4th root of $f(x)$ be $m$, then $f(x)=(x-1)(x-2)(x-3)(x-m)$, so $f(0)+f(4)=6 m+6(4-m)=24$.