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?
 A: 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 $$P(12) + P(-8) = 17940 + 990a + 1900 - 990a = 19840.$$
A: Other way doing this:
We try to find reals $e,f,g$ such that $P(12)+P(-8)=eP(1)+fP(2)+gP(3)$. So, if we try to equal "$x^k$ evaluated", we gain a system of equations:
$$
\left\{\begin{array}{ccc}
1^ke+2^kf+3^kg&=&12^k+(-8)^k
\end{array}\right.,\quad k=0,\cdots,4
$$
In particular,
$$
\left\{\begin{array}{ccc}
e+f+g&=&1+1\\
e+2f+3g&=&12+(-8)\\
e+2^2f+3^2g&=&12^2+(-8)^2
\end{array}\right.
$$
and we obtain $e=100,f=-198,g=100$. Verifying the others values for $k$:
$$
\left\{\begin{array}{ccc}
100+2^3(-198)+3^3\cdot100&=&12^3+(-8)^3\\
100+2^4\cdot(-198)+3^4\cdot100&=&12^4+(-8)^4 + 19800
\end{array}\right.
$$
So, $P(12)+P(-8)=100P(1)-198P(2)+100P(3)+1980=19840$
A: Two remarks, to avoid almost every computation:


*

*The polynomial $P(x)-10x$ has roots $1$, $2$ and $3$, hence there exists a polynomial $Q$ such that $P(x)-10x=(x-1)(x-2)(x-3)Q(x)$. 

*The polynomial $P(x)-10x$ has degree $4$ and leading coefficient $1$, hence $Q(x)=x+z$ for some unknown constant $z$ whose value will be irrelevant.


Thus, $P(12)+P(-8)=10\cdot(12-8)+11\cdot10\cdot9\cdot(12+z)+9\cdot10\cdot11\cdot(8-z)$, that is, $P(12)+P(-8)=10\cdot4+11\cdot10\cdot9\cdot(12+z+8-z)=40+990\cdot20=19840$.
