Đáp án:
$\begin{array}{l}
{x^4} - 3{x^3} + 4{x^2} - 3 - x\\
= {x^4} - 2{x^3} + 3{x^2} - {x^3} + 2{x^2} - 3x\\
- {x^2} + 2x - 3\\
= {x^2}\left( {{x^2} - 2x + 3} \right) - x.\left( {{x^2} - 2x + 3} \right)\\
- \left( {{x^2} - 2x + 3} \right)\\
= \left( {{x^2} - 2x + 3} \right)\left( {{x^2} - x - 1} \right)\\
\Leftrightarrow \left( {{x^4} - 3{x^3} + 4{x^2} - 3 - x} \right):\left( {{x^2} - 2x + 3} \right)\\
= \left( {{x^2} - 2x + 3} \right)\left( {{x^2} - x - 1} \right):\left( {{x^2} - 2x + 3} \right)\\
= {x^2} - x - 1\\
b)\\
14x - 13{x^2} - 15 + 6{x^3}\\
= 6{x^3} - 13{x^2} + 14x - 15\\
= 6{x^3} - 3{x^2} + 9x - 10{x^2} + 5x - 15\\
= 3x\left( {2{x^2} - x + 3} \right) - 5\left( {2{x^2} - x + 3} \right)\\
= \left( {2{x^2} - x + 3} \right)\left( {3x - 5} \right)\\
\Leftrightarrow \left( {14x - 13{x^2} - 15 + 6{x^3}} \right):\left( { - x + 2{x^2} + 3} \right)\\
= \left( {2{x^2} - x + 3} \right)\left( {3x - 5} \right):\left( {2{x^2} - x + 3} \right)\\
= 3x - 5
\end{array}$
