$T(n) = 4T({n/2}) + \theta(n\log{n})$ using Master Theorem

Your recurrence relation falls into Case 1: $f(n) = n \log n$ is $O(n^{log_{b}{a}-\epsilon}) = O(n^{2-\epsilon})$.

To show why this is Case 1, as Louis says, logarithmic functions ($\log n$) are asymptotically bounded by polynomial functions ($n^a$, where $a > 0$). This can be shown by taking the limit: $$ \lim_{n \to \infty} \frac{\log n}{n^a} = 0 $$ through L'Hôpital's rule. In particular, $\log n \in O(n^{1-\epsilon})$ for small $\epsilon$. (We can go even further and say that $\log n \in o(n^{1-\epsilon})$.)

Then by multiplying both sides by $n$, (an allowed operation in big-O notation), $n \log n \in O(n^{2-\epsilon})$.

Therefore by the Master Theorem, $T(n)$ is $\Theta(n^2)$.