2022-II-09

06-07-2023 app.cch 尚無留言

答案：D

由於 $g (x)$ 可被 $x + 2 a$ 整除，可得

$\begin{array}{rcl} g (- 2 a) & = & 0 \\ (- 2 a)^{2} + a (- 2 a) + b & = & 0 \\ 4 a^{2} - 2 a^{2} + b & = & 0 \\ b & = & - 2 a^{2} \end{array}$

所求的餘式

$\begin{array}{cl} = & g (2 a) \\ = & (2 a)^{2} + a (2 a) + b \\ = & 6 a^{2} + (- 2 a^{2}) \\ = & 4 a^{2} \end{array}$