<s>[INST] Calculate a quartic monic polynomial given that $P(1)=10,P(2)=20$ and $P(3)=30$. Let 
$$P(x)=x^4+ax^3+bx^2+cx+d$$
where $a,b,c$ and $d$ are constants.
If $P(1)=10,P(2)=20$ and $P(3)=30$, then what is the value of $$\frac{P(12)+P(-8)}{10}$$
I tried substituting $x=1,2,3$ in $P(x)$ to get three linear equations in $a,b,c$ and $d$, but it ended up being quite difficult. 
$$a+b+c+d=9$$
$$8a+4b+2c+d=4$$
$$27a+9b+3c+d=-51$$
Is there a simpler method to this question?
 [/INST] Define $Q(x) = P(x) - 10x$, then$$
P(1) = 10,\ P(2) = 20,\ P(3) = 30 \Longrightarrow Q(1) = Q(2) = Q(3) = 0.
$$
Since $\deg Q = 4$ and $Q$ is monic, there exists $x_0 \in \mathbb{R}$ such that$$
Q(x) = (x - 1)(x - 2)(x - 3)(x - x_0).
$$
Therefore,\begin{align*}
&\mathrel{\phantom{=}}{} P(12) + P(-8)\\
&= Q(12) + Q(-8) + 40\\
&= 11 × 10 × 9 × (12 - x_0) + (-9) × (-10) × (-11) × (-8 - x_0) +40\\
&= 11 × 10 × 9 × (12 - x_0) + 11 × 10 × 9 × (8 + x_0) +40\\
&= 11 × 10 × 9 × ((12 - x_0) + (8 + x_0)) +40\\
&= 11 × 10 × 9 × 20 +40\\
&=19840.
\end{align*}
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