Đáp án:
$\begin{array}{l}
a){x^3} - 5{x^2} + 8x - 4\\
= {x^3} - 2{x^2} - 3{x^2} + 6x + 2x - 4\\
= {x^2}\left( {x - 2} \right) - 3x\left( {x - 2} \right) + 2\left( {x - 2} \right)\\
= \left( {x - 2} \right)\left( {{x^2} - 3x + 2} \right)\\
= \left( {x - 2} \right).\left( {{x^2} - 2x - x + 2} \right)\\
= \left( {x - 2} \right)\left( {x - 2} \right)\left( {x - 1} \right)\\
= {\left( {x - 2} \right)^2}.\left( {x - 1} \right)\\
b)4{x^2} - 25 - \left( {2x - 5} \right)\left( {2x + 3} \right)\\
= \left( {2x - 5} \right)\left( {2x + 5} \right) - \left( {2x - 5} \right)\left( {2x + 3} \right)\\
= \left( {2x - 5} \right)\left( {2x + 5 - 2x - 3} \right)\\
= 2\left( {2x + 5} \right)\\
c){\left( {{x^2} + x} \right)^2} + 4{x^2} + 4x - 12\\
= {\left( {{x^2} + x} \right)^2} + 4\left( {{x^2} + x} \right) - 12\\
Đặt:\left( {{x^2} + x} \right) = a\\
\Leftrightarrow {a^2} + 4a - 12\\
= {a^2} + 6a - 2a - 12\\
= \left( {a + 6} \right)\left( {a - 2} \right)\\
= \left( {{x^2} + x + 6} \right)\left( {{x^2} + x - 2} \right)\\
= \left( {{x^2} + x + 6} \right)\left( {x - 1} \right)\left( {x + 2} \right)\\
e) - 14{x^2} + 39x - 10\\
= - 14{x^2} + 4x + 35x - 10\\
= - 2x\left( {7x - 2} \right) + 5\left( {7x - 2} \right)\\
= \left( {7x - 2} \right)\left( { - 2x + 5} \right)
\end{array}$
