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.
 A: We know that 
$$1+a+b+c+d=10$$
$$16+8a+4b+2c+d=20$$
$$81+27a+9b+3c+d=30$$
And we want to know
$$f(12)+f(-8)=24832+1216a+208b+4c+2d$$
Knowing that we can add the 24,832 later, it is sufficient to know 
$$608a+104b+2c+d$$ 

We want $z$ times the first equation, $y$ times the second  and $x$ times the third. This gives the equations, ignoring the number of $a$s:
$$x+y+z=1$$
$$3x+2y+z=2$$
$$9x+4y+z=104$$
First two equations give $2x+y=1$, first and second gives $8x+3y=103$, so $2x=100$ and $x=50$. This gives $y=-99$ and $z=50$. 
You know why I am happy? Because it turns out that $50\cdot 27-99\cdot8+50\cdot1=608$.

So 50 times the first equation plus -99 times the second plus 50 times the third gives $608a+104b+2c+d-2516=20$, thus $24832+1216a+208b+4c+2d=19840$
A: 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, 
$$
40 = g(-1) + g(1) = 2\cdot 1^4 + 2B\cdot 1^2 + 2D = 2 + 2B + 2D\\
40 = 2g(0) = g(0) + g(-0) = 2\cdot 0^4 + 2B\cdot 0^2 + 2D = 2D\\
g(10) + g(-10) = 2\cdot 10^4 + 2B\cdot 10^2 + 2D = 20000 - 200 + 40 = 19840
$$
A: 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
$$\begin{align}
f(12) + f(-8) 
&= 10(12 - 8) + 990(12-t) - 990(-8-t)\\
&= 40 + 990 \times 20 = 19840
\end{align}$$
A: HINT: Solve the system
$$1+a+b+c+d=10$$
$$16+8a+4b+2c+d=20$$
$$81+27a+9b+3c+d=30$$
