The trick is to simplify it out to a certain point, at which it becomes fairly simple as you'll see:

To begin simplifying, the first step is

\(128 + [(\frac{Q}{2})-4)*(-\frac{Q}{2}+4)]^z\)

Now we take out the -1 on the third part to get two identical parts which are multiplied, which we can combine together by squaring

\(128 - ((\frac{Q}{2}-4)^2)^z\)

Now it becomes a relatively simple way to find out what the maximum value is for the equation.

we want to find out what Q is when \((\frac{Q}{2}-4) = 0\)... this is because we know that the exponent is going to be even (since it's multiplied by 2), so that no matter what it will be possitive. Since we're subtracting from 128 a number that has to be positive, we want the lowest positive number possible in order to have the maximum.

if you solve for \((\frac{Q}{2}-4) = 0\) you get that \(Q = 8\)

So choice D.