Ans: A
$\begin{array}{rcl}
(x+3)^2+mx & \equiv & (x-n)(x+1)+2n \\
x^2+6x+9 +mx & \equiv & x^2 +x -nx-n+2n \\
x^2+(6+m)x +9 & \equiv & x^2 +(1-n)x +n
\end{array}$
By comparing the coefficients, we have
$\left\{ \begin{array}{ll}
6+m =1-n & \ldots \unicode{x2460} \\
9 = n & \ldots \unicode{x2461}
\end{array}\right.$
Sub. $\unicode{x2461}$ into $\unicode{x2460}$, we have
$\begin{array}{rcl}
6+m & = & 1-(9) \\
m & = & -14
\end{array}$
