Xét \(Q\left(x\right)=P\left(x\right)-10x\)
Có \(Q\left(1\right)=P\left(1\right)-10=10-10=0\)
\(Q\left(2\right)=P\left(2\right)-2.10=0\) ; \(Q\left(3\right)=P\left(3\right)-3.10=0\)
\(\Rightarrow Q\left(x\right)\) luôn có ít nhất 3 nghiệm \(x=\left\{1;2;3\right\}\)
\(\Rightarrow Q\left(x\right)=\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-x_0\right)\) với \(x_0\) là số thực bất kì
\(\Rightarrow P\left(x\right)=Q\left(x\right)+10x=\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-x_0\right)+10x\)
\(\Rightarrow P\left(12\right)=\left(12-1\right)\left(12-2\right)\left(12-3\right)\left(12-x_0\right)+10.12=12000-990x_0\)
\(P\left(-8\right)=\left(-8-1\right)\left(-8-2\right)\left(-8-3\right)\left(-8-x_0\right)-10.8=7840+990x_0\)
\(\Rightarrow P\left(12\right)+P\left(-8\right)=12000-990x_0+7890+990x_0=19840\)