So I have this problem where I need to find $E(4X+3Y-2Z^2-W^2+8)$ where $W,X,Y,Z$ are all standard normal and I'm kind of confused on how to find the expected value here. I thought to do it we just had to add the means together like this: $E(X_1+X_2+X_3)=\mu_1+\mu_2+\mu_3$ as long as they are iid of course. How come the answer is $5$?
I tried doing $\begin{align}E(4X+3Y-2Z^2-W^2+8)\\ =E(4X)+E(3Y)-E(2Z^2)-E(W^2)+E(8)\\ =4E(X)+3E(Y)-2E(Z^2)-E(W^2)+8\\=(4*0)+(3*0)-(2*0)-0+8\\=8\end{align}$ And quite obviously $8$ doesn't equal $5$. So I'm just not sure where I went wrong.