Ans: A
$\begin{array}{rcl}
(x+2)(x+c)+12 & \equiv & x(x+d)+6c(x+1) \\
x^2+cx+2x+2c+12 & \equiv & x^2 +dx+6cx+6c \\
x^2 +(c+2)x+(2c+12) & \equiv & x^2+(d+6c)x+6c
\end{array}$
By comparing the coefficients of both sides, we have
$\left\{ \begin{array}{ll}
c+2=d+6c & \ldots \unicode{x2460} \\
2c+12=6c & \ldots \unicode{x2461}
\end{array}\right.$
From $\unicode{x2461}$, we have
$\begin{array}{rcl}
2c+12 & = & 6c \\
4c & = & 12 \\
c & = & 3
\end{array}$
Sub. $c=3$ into $\unicode{x2460}$, we have
$\begin{array}{rcl}
3+2 & = & d+6(3) \\
5 & = & d+18 \\
d & = & -13
\end{array}$
