答案：B
$\begin{array}{rcl}
px(x-1)+x^2 & \equiv & qx(x-2)+4x \\
px^2 – px +x^2 & \equiv & qx^2 -2qx +4x \\
(p+1)x^2 – px & \equiv & qx^2 + (-2q +4) x
\end{array}$
$\begin{array}{rcl}
px(x-1)+x^2 & \equiv & qx(x-2)+4x \\
px^2 – px +x^2 & \equiv & qx^2 -2qx +4x \\
(p+1)x^2 – px & \equiv & qx^2 + (-2q +4) x
\end{array}$
透過比較兩方的係數，可得
$\left\{ \begin{array}{ll}
p+1 = q & \ldots \unicode{x2460} \\
-p = -2q+4 & \ldots \unicode{x2461}
\end{array} \right.$
把 $\unicode{x2460}$ 代入 $\unicode{x2461}$，可得
$\begin{array}{rcl}
-p & = & -2(p+1) +4 \\
-p & = & -2p -2 +4 \\
p & = & 2
\end{array}$
