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.

The fact that there are three equiations for four unknowns states that the solution will have one unconstrained unknown. Since you're asked for a concrete answer, the $f(12) + f(-8)$ is a specially crafted expression that doesn't depend that particular unconstrained unknown.
Let's simplifiy things by introducing $y = x - 2$. Let $g(y)$ be $f(x)$ for this new variable: 
$$
g(y) = f(y + 2) = (y + 2)^4 + a (y+2)^3 + \dots = 
y^4 + A y^3 + B y^2 + C y + D.
$$
$g(y)$ has similar constraints, 
$$
g(-1) = f(1) = 10\\
g(0) = f(2) = 20\\
g(1) = f(3) = 30
$$
and we're asked about $f(12) + f(-8) = g(10) + g(-10)$. Note the symmetry in constraints and the value we are interesting in.
Now, 
$$
40 = g(-1) + g(1) = 2\cdot 1^4 + 2B\cdot 1^2 + 2D = 2 + 2B + 2D\\
40 = 2g(0) = g(0) + g(-0) = 2\cdot 0^4 + 2B\cdot 0^2 + 2D = 2D\\
g(10) + g(-10) = 2\cdot 10^4 + 2B\cdot 10^2 + 2D = 20000 - 200 + 40 = 19840
$$
