4. $f(x)=x^{4}+a x^{3}+b x^{2}+c x+d$, here $a, b, c, d$ are real numbers. It is known that $f(1)=5, f(2)=10$, $f(3)=15$, then $f(8)+f(-4)$ is $(\quad)$.
(A) 2500
(B) uncertain
(C) 2540
(D) 860
4. (C).
$$
\begin{array}{c}
f(x)-5 x \\
=(x-1)(x-2)(x-3)(x+h), \\
f(8)-40=210(8+h), \\
f(-4)+20=-210(h-4), \\
\text { then, } f(8)+f(-4)=2540 .
\end{array}
$$