It is well-known that in a model category $M$ (assuming functorial factorizations as part of the axioms) we can replace every object with a cofibrant one, up to equivalence. Indeed, we can find a natural transformation $q \ \colon Q\Rightarrow \text{id}_M $ which is pointwise a trivial fibration and $QX$ is cofibrant for every $X$.
Suppose that $X$ is already cofibrant. In that case we can find $s_X \ \colon X\to QX$ s.t. $q_Xs_X=\text{id}_X$ by the RLP of the trivial fibration $q_X$ with respect to the cofibration $\emptyset\to X$. I'm trying to show that $s_X$ is an actual inverse (making $q_X$ an isomorphism), but I'm stuck with the proof that $s_Xq_X=\text{id}_{QX}$.
This is usually not true. For instance, in the typical examples where the functorial factorizations are constructed by a small object argument, the construction pays no regard to whether the map you started with was actually already a cofibration and will massively change your object in almost all cases. For instance, for the usual model structure on topological spaces, $QX$ will take a space and "formally" turn it into a cell complex by repeatedly adjoining cells to solve every possible lifting problem with respect to the maps $S^{n-1}\to D^n$. Unless $X$ is empty, this will give a space which is much bigger than $X$ itself (for instance, for every point of $X$, you will end up adjoining cells along the constant maps from each $S^{n-1}$ to that point infinitely many times!).