<s>[INST] Finding a,b,c,d in a quartic expression Let $p(x)=x^4+ax^3+bx^2+cx+d$ where a,b,c,d are constants. If $p(1)=10$, $p(2)=20$, $p(3)=30$, compute $\frac {p(12)+p(-8)}{10}$. I have tried so far. 
\begin{align}
a+b+c+d=&9\\8a+4b+2c+d=&4\\27a+9b+3c+d=&-51 
\end{align}
Manipulating these, I got $6a+b=-25$. Now, $$\frac {p(12)+p(-8)}{10}=\frac{24832+1216a+208b+4c+2d}{10}$$ $$=\frac{24832+202(6a+b)+(4a+4b+4c)+2b+2d}{10}$$ $$=\frac{19782+(36-4d)+2b+2d}{10}$$ $$=\frac{19818+2b-2d}{10}$$ How do I get rid of the $2b-2d$?
 [/INST] (2012's answer is correct, but the algebraic manipulations doesn't reveal what is happening. This answer explains why we could calculate the expression, despite not having enough information.)
By the remainder factor theorem, since $p(x) - 10x$ is a monic quartic polynomial with roots 1, 2, and 3, hence
$$p(x) - 10x = ( x-1) (x-2) ( x-3) (x-k), $$
where $k$ is some constant.
Hence, $ p(12) - 120 = 11 \times 10 \times 9 \times (12-k)$ and $p(-8) - (-80) = (-9) \times (-10) \times (-11) \times (-8-k)$.
Note that the coefficients of $ (12-k) $ and $-(-8-k)$ are the same, namely $11\times 10 \times 9$, so we can add them up to get:
$$p(12) + p(-8) = 120 + (-80) + 11 \times 10 \times 9 \times (12+8) = 19840.$$
If you want a similar problem to practice what you learnt in this problem, try this math problem on Brilliant.
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