What would the Big Theta of this be? So I have this function $$2^n(n+2)^2\log_2n$$
And I simplified it down to this$$2^nn^2\log_2n + 2^{2+n}n\log_2n + 2^{2+n}\log_2n$$
However, I'm not sure what the Big Theta of this function would be. I'm thinking it's $\Theta(2^nn^2\log_2n)$, but I'm not sure if this is correct or if I should simplify it. Could anyone help.

There isn't a unique answer (e.g., $\Theta(n)$ and $\Theta(n+2)$ are the same), but your answer is valid since
$$\frac{2^n (n+2)^2 \log_2 n}{2^n n^2 \log_2 n} = \left(1+\frac{2}{n} \right)^2 \to 1.$$
I don't think there is another simpler form you could use.
