Ans: A
$\begin{array}{rcl}
4x^2 + m(x+1) + 28 & \equiv & mx(x+3) + n(x-4) \\
4x^2 + mx + (28+m) & \equiv & mx^2 + (3m + n)x – 4n
\end{array}$
$\begin{array}{rcl}
4x^2 + m(x+1) + 28 & \equiv & mx(x+3) + n(x-4) \\
4x^2 + mx + (28+m) & \equiv & mx^2 + (3m + n)x – 4n
\end{array}$
By comparing the coefficients of both sides, we have
$\left\{ \begin{array}{ll}
4 = m & \ldots \unicode{x2460} \\
m = 3m+ n & \ldots \unicode{x2461} \\
28 + m = -4n & \ldots \unicode{x2462}
\end{array} \right.$
Sub. $\unicode{x2460}$ into $\unicode{x2462}$, we have
$\begin{array}{rcl}
28 + (4) & = & -4n \\
-4n & = & 32 \\
n & = & -8
\end{array}$
