To determine if a relation is BCNF we examine it's functional dependencies.
It is in BCNF if for each FD X→Y, we either have
- X→Y is a trivial functional dependency (Y ⊆ X)
- X is a superkey for schema R.
The FDs are A→C, D→CB, AC→E. Let's start with the first FD A→C.
A→C is not trivial because C ∉ A. A→A is trivial dependency for instance.
Now is A→C a superkey? To check that we compute the closure of left hand side of the FD, in this case A. The closure is all elements logically implied by A. [A]+ = A ∪ C ∪ E = ACE or so we have A→ACE.
ACE is not a superkey, because it does contain all attributes of the relation.
So the relation is not in BCNF, because A→C is neither trivial or a superkey.
There are other violations of BCNF too. [D]+ = BCD which is not a superkey or trivial. [AC]+ = ACE which is not a superkey or trivial.
Hope this helps! I think everything is correct but I'm studying for finals right now and trying to learn a lot of this material as well.