If $3x^3-9x^2+kx-12$ is divisible by $x-3$, then it is also divisible by:
$\textbf{(A)}\ 3x^2-x+4\qquad \textbf{(B)}\ 3x^2-4\qquad \textbf{(C)}\ 3x^2+4\qquad \textbf{(D)}\ 3x-4 \qquad \textbf{(E)}\ 3x+4$
If $3x^3-9x^2+kx-12$ is divisible by $x-3$, then by the [Remainder Theorem](https://artofproblemsolving.com/wiki/index.php/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 $\boxed{\textbf{(C)}}$.