Ans: D
$\begin{array}{rcl}
\mbox{LHS} & = & m(x-3)^2+n(x+1)^2 \\
& = & m(x^2-6x+9)+n(x^2+2x+1) \\
& = & (m+n)x^2 +(-6m+2n)x +(9m+n)
\end{array}$
$\begin{array}{rcl}
\mbox{LHS} & = & m(x-3)^2+n(x+1)^2 \\
& = & m(x^2-6x+9)+n(x^2+2x+1) \\
& = & (m+n)x^2 +(-6m+2n)x +(9m+n)
\end{array}$
By comparing the coefficients of both sides, we have
$\left\{ \begin{array}{ll}
m+n=1 & \dots \unicode{x2460}\\
-6m+2n=-38 & \ldots\unicode{x2461} \\
9m+n =41 & \ldots \unicode{x2462}
\end{array}\right. $
$\unicode{x2462} – \unicode{x2460}$, we have
$\begin{array}{rcl}
8m & = & 40 \\
m & = & 5
\end{array}$
