Using Theta Notation $\Theta$ how can I prove that $(13n + 3)(9n + 1)(\log(4n^2 + 100))$ is an element of $\Theta(n^2 \log n)$ I'm having a hard time figuring out what systematic approach I need to follow to solve questions like these :/ Do I expand and use the Max() principle to show that it reduces to the RHS?
Please help :(
 A: So you would like to show 
$$f(n) = \Theta(n^2 \log n)$$
This means that we have to find $c_1,c_2,n_0$ for every $n \geq n_0$, we have
\begin{equation}
 c_1 n^2 \log n \leq f(n) \leq c_2 n^2 \log n
\end{equation}
where 
\begin{equation}
 f(n) = (13n + 3)(9n + 1)(\log(4n^2 + 100))
\end{equation}
Notice that you can write
\begin{equation}
 f(n) =(117n^2 + 40n + 3)\log(4n^2 + 100) 
\end{equation}
Now it is clear that for $c_1 = 1$, we have 
\begin{equation}
 f(n) \geq n^2 \log n
\end{equation}
This leave us to find $c_2$ and $n_0$. Assuming the $\log$ is of base $2$ (even though it could be generalized for any base), we have
\begin{equation}
 \begin{split}
  f(n) &=(117n^2 + 40n + 3)\log(4n^2 + 100)  \\
  &\leq (117n^2 + 117n^2 + 117n^2)\log(4n^2 + 100) \\
  &= 3(117n^2) \log(4n^2 + 100)\\
  &\leq 3(117n^2) \log(4n^2 + 4n^2)\\
  &= 351n^2 \log (8n^2)\\
  &= 351n^2 \log (8) + 351n^2\log(n^2)\\
  &= 3(351n^2) + 702n^2\log(n) \\
  &= 1053n^2 + 702n^2\log(n) \\
  &\leq 1053n^2 + 1053n^2\log(n) \\
  &\leq 1053n^2\log(n) + 1053n^2\log(n) \\
  &= 2016n^2\log(n) \\
  &= c_2 n^2\log(n)
 \end{split}
\end{equation}
So choosing $c_2 = 2016$ for $n_0 = 2$, we get that $f(n) \leq c_2 n^2\log(n)$
