Giải thích các bước giải:
Ta có:
\[\begin{array}{l}
{x^3} - 4{x^2} + 5x - 1\\
= \left( {{x^3} - 3{x^2}} \right) - \left( {{x^2} - 3x} \right) + \left( {2x - 6} \right) + 5\\
= {x^2}\left( {x - 3} \right) - x\left( {x - 3} \right) + 2\left( {x - 3} \right) + 5\\
= \left( {x - 3} \right)\left( {{x^2} - x + 2} \right) + 5\\
\Rightarrow \left( {{x^3} - 4{x^2} + 5x - 1} \right) \vdots \left( {x - 3} \right) \Leftrightarrow 5 \vdots \left( {x - 3} \right)\\
\Rightarrow \left( {x - 3} \right) \in U\left( 5 \right) = \left\{ { \pm 1; \pm 5} \right\}\\
b;\\
10{x^2} - x + 10\\
= \left( {10{x^2} - 10x} \right) + \left( {9x - 9} \right) + 19\\
= 10x\left( {x - 1} \right) + 9\left( {x - 1} \right) + 19\\
= \left( {10x + 9} \right)\left( {x - 1} \right) + 19\\
\Rightarrow 19 \vdots \left( {x - 1} \right)\\
c;\\
{x^4} - 5{x^3} - 3{x^2} + 17x - 17\\
= \left( {{x^4} - 5{x^3}} \right) - \left( {3{x^2} - 15x} \right) + \left( {2x - 10} \right) - 7\\
= {x^3}\left( {x - 5} \right) - 3x\left( {x - 5} \right) + 2\left( {x - 5} \right) - 7\\
= \left( {x - 5} \right)\left( {{x^3} - 3x + 2} \right) - 7\\
\Rightarrow 7 \vdots \left( {x - 5} \right)
\end{array}\]
