<s>[INST] Value of $P(12)+P(-8)$ if $P(x)=x^{4}+ax^{3}+bx^{2}+cx+d$, $P(1)=10$, $P(2)=20$, $P(3)=30$ 
What will be the value of $P(12)+P(-8)$ if $P(x)=x^{4}+ax^{3}+bx^{2}+cx+d$
  provided that $P(1)=10$, $P(2)=20$, $P(3)=30$?

I put these values and got three simultaneous equations in $a, b, c, d$. What is the smarter way to approach these problems?
 [/INST] I'm not sure this is the smartest way, but here's one way.
Define $Q(x) = P(x) - x^4$ which satisfies the conditions $$Q(1) = 9, Q(2) = 4, Q(3) = -51.$$
So $a,b,c,d$ satisfy the matrix identity
$$\begin{pmatrix} 1 & 1 & 1 & 1 \\ 1 & 2 & 4 & 8 \\ 1 & 3 & 9 & 27 \end{pmatrix} \begin{pmatrix} d \\ c \\ b \\ a \end{pmatrix} = \begin{pmatrix} 9 \\ 4 \\ -51 \end{pmatrix}.$$  To solve this, find the kernel of the matrix (a bit of linear algebra), which turns out to be $$<\begin{pmatrix} -6 \\ 11 \\ -6 \\ 1 \end{pmatrix}>$$ and add a particular solution, for example where $a = 0$: $b = -25, c = 70, d = -36$.
So your polynomial is given by $$P(x) = x^4 + ax^3 + (-25-6a)x^2 + (11 + 70a)x + (-36 - 6a)$$ for some $a$ which we can't determine.
You then get $$P(12) + P(-8) = 17940 + 990a + 1900 - 990a = 19840.$$
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