答案：B
由於 $f(x)$ 可被 $2x+1$ 整除，可得
由於 $f(x)$ 可被 $2x+1$ 整除，可得
$\begin{array}{rcl}
f(\dfrac{-1}{2}) & = & 0 \\
4(\dfrac{-1}{2})^3 +k(\dfrac{-1}{2}) + 3 & = & 0 \\
\dfrac{-1}{2} – \dfrac{1}{2} k + 3 & = & 0 \\
k & = & 5
\end{array}$
所求的餘數
$\begin{array}{cl}
= & f(-1) \\
= & 4(-1)^3 +5(-1) + 3 \\
= & -6
\end{array}$
