Đáp án:
$\begin{array}{l}
a){x^4} - 2{x^2} - 3\\
= {x^4} - 3{x^2} + {x^2} - 3\\
= \left( {{x^2} - 3} \right)\left( {{x^2} + 1} \right)\\
b){\left( {x + 1} \right)^4} - 2{\left( {{x^2} + 2x + 1} \right)^2} + 1\\
= {\left( {x + 1} \right)^4} - 2{\left( {x + 1} \right)^4} + 1\\
= 1 - {\left( {x + 1} \right)^4}\\
= 1 - {\left( {{x^2} + 2x + 1} \right)^2}\\
= \left( {1 - {x^2} - 2x - 1} \right)\left( {1 + {x^2} + 2x + 1} \right)\\
= - x\left( {x + 2} \right).\left( {{x^2} + 2x + 2} \right)\\
c)\left( {x + 1} \right)\left( {x + 5} \right)\left( {x + 2} \right)\left( {x + 4} \right) - 4\\
= \left( {{x^2} + 6x + 5} \right)\left( {{x^2} + 6x + 8} \right) - 4\\
Đặt:\left( {{x^2} + 6x + 5} \right) = a\\
\Leftrightarrow a.\left( {a + 3} \right) - 4\\
= {a^2} + 3a - 4\\
= {a^2} + 4a - a - 4\\
= \left( {a + 4} \right)\left( {a - 1} \right)\\
= \left( {{x^2} + 6x + 5 + 4} \right)\left( {{x^2} + 6x + 5 - 1} \right)\\
= \left( {{x^2} + 6x + 9} \right)\left( {{x^2} + 6x + 4} \right)\\
= {\left( {x + 3} \right)^2}\left( {{x^2} + 6x + 4} \right)\\
d){\left( {{x^2} + 2x - 1} \right)^2} - 3x\left( {{x^2} + 2x - 1} \right) + 2{x^2}\\
Đặt:{x^2} + 2x - 1 = a\\
\Leftrightarrow {a^2} - 3x.a + 2{x^2}\\
= {a^2} - 2x.a - x.a + 2{x^2}\\
= \left( {a - 2x} \right)\left( {a - x} \right)\\
= \left( {{x^2} + 2x - 1 - 2x} \right)\left( {{x^2} + 2x - 1 - x} \right)\\
= \left( {{x^2} - 1} \right)\left( {{x^2} + x - 1} \right)\\
= \left( {x - 1} \right)\left( {x + 1} \right)\left( {{x^2} + x - 1} \right)
\end{array}$
