<s>[INST] 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.
 [/INST] 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, 
$$
40 = g(-1) + g(1) = 2\cdot 1^4 + 2B\cdot 1^2 + 2D = 2 + 2B + 2D\\
40 = 2g(0) = g(0) + g(-0) = 2\cdot 0^4 + 2B\cdot 0^2 + 2D = 2D\\
g(10) + g(-10) = 2\cdot 10^4 + 2B\cdot 10^2 + 2D = 20000 - 200 + 40 = 19840
$$
</s>