smc

If $3x^3-9x^2+kx-12$ is divisible by $x-3$, then by the Remainder Theorem, plugging in $3$ in the cubic results in $0$. $$3\cdot 3^3-9\cdot 3^2+3k-12=0$$ Combine like terms to get $$3k-12=0$$ Thus, $k=4$. The cubic is $3x^3-9x^2+4x-12$, and it can be factored (by grouping or synthetic division) into $$(3x^2+4)(x-3)$$ Thus, the answer is $\boxed{3x^2+4}$.