2022-II-03

$\begin{array}{rcl}(x+3{)}^2+mx& \equiv & (x-n)(x+1)+2n\\ x^2+6x+9+mx& \equiv & x^2+x-nx-n+2n\\ x^2+(6+m)x+9& \equiv & x^2+(1-n)x+n\end{array}$

透過比較係數，可得

$\{\begin{array}{ll}6+m=1-n& \dots \text{①}\\ 9=n& \dots \text{②}\end{array}$

把 $\text{②}$ 代入 $\text{①}$，可得

$\begin{array}{rcl}6+m& =& 1-(9)\\ m& =& -14\end{array}$