\(P\left(x\right)=x^4+ax^3+bx^2+cx+d\)
Đặt \(Q\left(x\right)=P\left(x\right)-10x\) \(\Rightarrow\left\{{}\begin{matrix}Q\left(1\right)=P\left(1\right)-10.1=10-10=0\\Q\left(2\right)=P\left(2\right)-10.2=20-20=0\\Q\left(3\right)=P\left(3\right)-10.3=30-30=0\end{matrix}\right.\)
\(\Rightarrow Q\left(x\right)\) có 3 nghiệm \(x=\left\{1;2;3\right\}\Rightarrow Q\left(x\right)=\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-a\right)\)
Mà \(Q\left(x\right)=P\left(x\right)-10x\Rightarrow P\left(x\right)=Q\left(x\right)+10x\)
\(\Rightarrow P\left(x\right)=\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-a\right)+10x\)
\(P\left(12\right)=990\left(12-a\right)+120=12000-990a\)
\(P\left(-8\right)=-990\left(-8-a\right)-80=990a+7840\)
\(\Rightarrow\frac{P\left(12\right)+P\left(-8\right)}{10}=\frac{12000-990a+990a+7840}{10}=1984\)
