Giải thích các bước giải:
a) Ta có:
$\begin{array}{l}
f\left( x \right) = {x^3} + {x^2} - x + m + 2\\
= x\left( {{x^2} - x + 1} \right) + 2\left( {{x^2} - x + 1} \right) + m\\
= \left( {x + 2} \right)\left( {{x^2} - x + 1} \right) + m\\
= \left( {x + 2} \right)g\left( x \right) + m
\end{array}$
Vậy $f\left( x \right) \vdots g\left( x \right)$
$\begin{array}{l}
\Leftrightarrow \left( {x + 2} \right)g\left( x \right) + m \vdots g\left( x \right)\\
\Leftrightarrow m \vdots g\left( x \right)\\
\Leftrightarrow m = 0
\end{array}$
Vậy $m=0$
b) Ta có:
$\begin{array}{l}
f\left( x \right) = {x^4} - {x^3} + 6{x^2} - x + m\\
= {x^2}\left( {{x^2} - x + 5} \right) + \left( {{x^2} - x + 5} \right) + m - 5\\
= \left( {{x^2} + 1} \right)\left( {{x^2} - x + 5} \right) + m - 5\\
= \left( {{x^2} + 1} \right)g\left( x \right) + m - 5
\end{array}$
Để $f\left( x \right) \vdots g\left( x \right)$
$\begin{array}{l}
\Leftrightarrow \left( {{x^2} + 1} \right)g\left( x \right) + m - 5 \vdots g\left( x \right)\\
\Leftrightarrow m - 5 \vdots g\left( x \right)\\
\Leftrightarrow m - 5 = 0\\
\Leftrightarrow m = 5
\end{array}$
Vậy $m=5$
