Is it true that $(2^n+n^2)(n^3+3^n)$ is $O(6^n)$?

Question

$(2^n+n^2)$ is $O(2^n)$ and $(n^3+3^n)$ is $O(3^n)$, therefore I conclude that $(2^n+n^2)(n^3+3^n)$ is $O(2^n*3^n)=O(6^n)$

Answer 1

Your answer is correct, yet if you were not 100% sure you can always check using the definition of "big O": $$\limsup_{n\rightarrow\infty}\frac{(2^n+n^2)(n^3+3^n)}{6^n}=\limsup_{n\rightarrow\infty}\frac{n^5+n^32^n+n^23^n+6^n}{6^n}=\limsup_{n\rightarrow\infty}(n^5(1/6)^n+n^3(1/3)^n+n^2(1/2)^n+1)=1$$

$$\Rightarrow (2^n+n^2)(n^3+3^n)=\mathcal{O }(6^n)$$

Answer 2

Yes ${}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}$