Compute $\frac{p(12)+p(-8)}{10}$ where $p(x)=x^4+ax^3+bx^2+cx+d$ I have got an olympiad problem which is as follow:

Compute $\frac{p(12)+p(-8)}{10}$ where $p(x)=x^4+ax^3+bx^2+cx+d$ and $p(1)=10$, $p(2)=20$, $p(3)=30$.

I have been told that answer is $1984$.
I thought applying the values and getting a relation between $a,b,c,d$ from $p(1), p(2), p(3)$ will suffice but the problem is that I will end up with with 4 variables and three equations.
I m a newbie to polynomials and I don't think I m gonna get the answer. So, please help me in this. Thanks.
 A: Let $q(x) = p(x)-10x$. Then, $q(x)$ is a monic polynomial, and $q(1) = q(2) = q(3) = 0$. 
So the roots of $q(x)$ are $x = 1, 2, 3$ and $r$ for some real number $r$.
Hence, we can write $q(x) = (x-1)(x-2)(x-3)(x-r)$.
Thus, $p(x) = (x-1)(x-2)(x-3)(x-r)+10x$ for some real number $r$. 
So, $p(12) = 11 \cdot 10 \cdot 9 \cdot (12-r) + 120 = 990(12-r)+120$, and $p(-8) = (-9) \cdot (-10) \cdot (-11) \cdot (-8-r) - 80 = -990(-8-r)-80$. 
Therefore, $\dfrac{p(12)+p(-8)}{10} = \dfrac{990(12-r)+120-990(-8-r)-80}{10} = 1984$.
A: We can write $p(x)-10x=0\forall x=1,2,3$
So using factor theorem $p(x)-10x = 0$ has three factors  $(x-1)\;,(x-2)\;,(x-3).$
Given $p(x)$ is a $4^{th}$ degree equation with leading coefficients $=1$
So let $x=r$ be the $4^{th}$ root of above equation
So $$p(x)-10x = (x-1)(x-2)(x-3)(x-r)$$
So $$p(12) = 11\cdot 10\cdot 9\cdot (12-r)+10\cdot 12$$
and $$p(-8) = 11\cdot 10 \cdot 9\cdot (8+r)-10\cdot 8$$
So $$\frac{p(12)+p(-8)}{10} = \frac{9\cdot 10 \cdot 11 \cdot 20+10\cdot 4}{10} = $$
