How to prove that ($56n^2+106n+48)(\log(264n^2+200)) = Θ(^2\log n)$
For all $n\geq 1$, $56n^2+106n+48&gt; 56n^2&gt; n^2$ and $\log (264n^2+200)&gt; \log 264n^2&gt;\log n$, so 
$$(56n^2+106n+48)\log(264n^2+200) &gt; n^2\log n\,,$$
i.e., you can take $c_1=1$.

Also for all $n\geq 1$, $56n^2+106n+48\leq 56n^2+106n^2+48n^2 = 210n^2$ and, for all $n\geq 200$, $264n+200 &lt; 265n$ so $\log(264n^2+200) &lt; \log 265n^2 = 2\log n + \log 265$. For all $n\ge 265$, $2\log n + \log 265\leq 3\log n$.  Therefore, for all $n\geq 265$,
$$(56n^2+106n+48)\log(264n^2+200) &lt; 630n^2\log n\,,$$
i.e., you can take $c_2=630$.