Solving $T(n) = 4T(n/2) + n^2log_2(n)$
When you applied the master theorem, f(n) is n^2 right. This is anyways asymptotically lesser than O(n) which is the n^2log(n). Hence at each node of the recursion tree, the O(n) would dominate computation hence T(n) is O(n). The reason why this is true is because when you apply the recursion tree the number of processes at each node if you calculate would be 2O(n/2)+4O(n/4)...... whose sum turns out to be lesser than O(n).