We start by rewriting \(\frac{16}{(2+x)^2}\) as \(16(2+x)^{-2}\).

Using the binomial expansion for \((1 + u)^n\), where \(u = \frac{x}{2}\) and \(n = -2\), we have:

\((1 + \frac{x}{2})^{-2} = 1 - 2\frac{x}{2} + \frac{(-2)(-3)}{2!} \left(\frac{x}{2}\right)^2 + \cdots\)

Simplifying, we get:

\(1 - x + \frac{3}{2}x^2 + \cdots\)

Now, multiply by 16:

\(16(1 - x + \frac{3}{2}x^2) = 16 - 16x + 24x^2\)

Thus, the expansion up to \(x^2\) is \(4 - 4x + 3x^2\).