Asymptotic upperbound in multiplication

Question

How can someone calculate the asymptotic upperbound of $2^nn^2$? The first term ($2^n$) grows much faster than the second, but saying that as a final result $2^nn^2 = O(2^n)$ would only be true in the case we had in addition, right? How does it work in multiplication?

Answer 1

We could say that for any positive $\epsilon$, we have $n^22^n=O(2^{n+\epsilon})$. That is occasionally useful.

You are right in saying that it is not $O(2^n)$.

Answer 2

It holds that $2^n n^2= O(2^n n^2)=O(2^n n^3)= \dots=O(2^n 2^m)=O(2^{n+m})$