Say $a,b$ are 1-forms, and $f$ is zero form, do we have $$fa\wedge b=a\wedge fb$$
I do not believe this is true as we have $$d(x\,dy\wedge dz)\neq d(dy\wedge x\,dz)$$ However, I just want to make sure this is the case.
Computation to show $d(x\,dy\wedge dz)\neq d(dy\wedge x\,dz)$:
Observe that $d(x\,dy\wedge dz)=d(x\,dy)\wedge dz+x\,dy\wedge d(dz)=(dx\wedge dy\wedge dz)+0=dx\wedge dy\wedge dz$
On the other hand, we have $d(dy\wedge x\,dz)=d(dy)\wedge x\,dz+dy\wedge (d(x\,dz))=dy\wedge dx\wedge dz$.
The LHS and RHS does not equal as we need to change a sign when we swap $dy$ and $dx$.
The zero forms should be passing through, I missed a negative sign when taking the derivative.