Given $f(1)=10,f(2)=20,f(3)=30$ find $f(12)+f(-8)$ for a 4-th degree monic polynomial If $f(x)=x^4+ax^3+bx^2+cx+d$.
Given $f(1)=10,f(2)=20,f(3)=30$ find $f(12)+f(-8)$.
This problem has troubled me a lot.The more I try to solve it,it becomes lengthier.
My problem is that there are four unknowns and only three equations.
Please help me.

Let $g(x) = f(x) - 10x$, we have 
$$g(1) = g(2) = g(3) = 0\quad\implies\quad (x-1)(x-2)(x-3) \;\;\text{ divides }\;\;g(x)$$
As a result, $f(x)$ has the form
$$f(x) = 10x + (x-t)(x-1)(x-2)(x-3)$$
for suitably chosen constant $t$. Notice 
$$\begin{align}
(12-1)(12-2)(12-3) &= 990\\
(-8-2)(-8-1)(-8-3) &= -990
\end{align}$$
we have
$$\begin{align}
f(12) + f(-8) 
&= 10(12 - 8) + 990(12-t) - 990(-8-t)\\
&= 40 + 990 \times 20 = 19840
\end{align}$$
