Ans: D
$\begin{array}{rcl}
(x-c)(x-4c) & = & (3c-x)(x-4c) \\
x^2 -cx -4cx +4c^2 & = & 3cx -x^2 -12c^2 +4cx \\
2x^2 -12cx +16c^2 & = & 0 \\
x^2 -6cx +8c^2 & = & 0 \\
(x-2c)(x-4c) & = & 0
\end{array}$
$\therefore x=2c$ or $x=4c$.
