\(\left\{{}\begin{matrix}P\left(1\right)=a+b+c+d=100\\P\left(-1\right)=-a+b-c+d=50\\P\left(0\right)=0+0+0+d=1\\P\left(2\right)=8a+4b+2c+d=120\end{matrix}\right.\) \(\begin{matrix}\left(1\right)\\\left(2\right)\\\left(3\right)\\\left(4\right)\end{matrix}\)
(3)=> d=1
(1) +(2) <=>2b+2 =150 => b=74
(1) -(2)<=>2(a+c) =50 ; a+c=25 ; c =25-a
(4) <=> \(8a+4.74+2\left(25-a\right)+1=120;a=\dfrac{-227}{6}\)
\(P\left(3\right)=27a+9b+3c+d=a+b+c+d+2\left(a+c\right)+24a+9b\)
\(P\left(3\right)=100+50-\dfrac{227}{6}.24+8\cdot74=-166\)