Is knowing ALL factors of a large number an NP?
NP is a class of decision problems. Only decision problems can belong to NP. The problem "factor a number" is not a decision problem.

You can make up many decision problems that attempt to capture factoring. Here is one: the language of all quartets $(n,i,j,b)$ such that the $j$th bit of the $i$th prime factor of $n$ is equal to $b$. This language is in NP, since the factorization of $n$ functions as a witness (this uses the fact that primality can be tested in polynomial time; in fact, it is enough to know that primality is in NP, which is an easier fact). It is strongly suspected that this language is not NP-complete.