Resolution on 3-SAT instance yields in polynomial many resolvents

A SAT instance in CNF with $n$ variables has at most $2^n$ resolvents, therefore the resolution method is not in polynomial time.

Considering a 3-SAT instance, we have at most $n^3 + n^2 + n$ many resolvents. Therefore, we can decide 3-SAT in polynomial time by resolution.

I am sure, my second statement is wrong, because otherwise we would have $P = NP$. But I can not see my mistake. Can someone bring clarification here?

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Your second statement is wrong, but I don't see how we can explain your mistake if you don't explain how you arrived at it. –  Eric Tressler Jul 4 '13 at 12:42
@EricTressler Roughly: All 1-Literal clauses: $n$ many. All 2-Literal clauses $n^2$ many. All 3-Literal clauses $n^3$ many. Calculate all clauses (polynomial many) and check for an empty clause. –  John Threepwood Jul 4 '13 at 12:49
The problem is that to determine if a CNF instance is satisfiable, you have to assign boolean values to each of the variables $x_i$ so that all of the existing clauses are satisfied simultaneously. You can't deal with them one by one, so you still have $2^n$ possible assignments of the $x_i$. –  Eric Tressler Jul 4 '13 at 13:38
@EricTressler Thank you for the reply. Resolution is refutation complete: A SAT instance is satisfiable iff the empty clause can not be derived as resolvent. So building all resolvents is sufficient to test for satisfiability. –  John Threepwood Jul 4 '13 at 13:50
There are only that many clauses but that only gives an upper bound for the time to test a single candidate solution. There are still $2^n$ candidates to test.