$\begin{array}{lcl} 6x^3\quad + \quad 7x^2\quad - \quad 4x\quad + \quad m^2 - 6m + 5 & \Big| & \underline{2x + 1\qquad}\\ \underline{6x^3\quad +\quad 3x^2}&\Big|& 3x^2 + 2x -1\\ \quad \quad \quad \quad \quad 4x^2\quad - \quad 4x & \Big| &\\ \quad \quad \quad \quad \quad \underline{4x^2\quad - \quad 2x}&\Big|&\\ \quad \quad \quad \quad \quad \quad \quad \quad - \quad 2x\quad + \quad m^2 - 6m + 5&\Big|&\\ \quad \quad \quad \quad \quad \quad \quad \quad \underline{- \quad 2x\quad -\quad 1\quad \quad \quad\quad \quad}&\Big|&\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad m^2 - 6m + 6&&\end{array}$
b) Ta có:
$A\quad \vdots \quad B \Leftrightarrow m^2 - 6m + 6 = 0$
$\Leftrightarrow m^2 - 6m + 9 = 3$
$\Leftrightarrow (m-3)^2 = 3$
$\Leftrightarrow \left[\begin{array}{l}m - 3 = \sqrt3\\m - 3 = -\sqrt3\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}m =3 + \sqrt3\\m =3 -\sqrt3\end{array}\right.$
