Is the complement of this decision problem in $P$?

Are there any two primes that are NOT a factor of $ M$ that multiply up to $ M$ ?

Fact: Any two primes that multiply up to $ M$ . Must be factors of $ M$ !

Thus because of the fact above an $ O(1)$ algorithm exists. It always outputs $ NO$

Complement

Are there any two primes that are a factor of $ M$ that multiply up to $ M$ ?

Fact: A complement of a decision problem does not always require to always return $ YES$ or $ NO$ . It can be either one!

(eg. $ M$ = 6 and two primes that multiply up to $ M$ are $ 3$ ,$ 2$ .)

Well, this I find interesting this is deciding $ Semi-Primes$ .

Question

Shouldn’t $ Semi-Primes$ be in $ P$ , because what was shown above?