Đáp án:
$\begin{array}{l}
a){x^2} + 5x - 6\\
= {x^2} + 6x - x - 6\\
= \left( {x + 6} \right)\left( {x - 1} \right)\\
b) - 6{x^2} + 7x - 2\\
= - 6{x^2} + 3x + 4x - 2\\
= \left( {2x - 1} \right)\left( { - 3x + 2} \right)\\
c){x^2} + 4x + 3\\
= {x^2} + 3x + x + 3\\
= \left( {x + 3} \right)\left( {x + 1} \right)\\
d)2{x^2} + 3x - 5\\
= 2{x^2} + 5x - 2x - 5\\
= \left( {2x + 5} \right)\left( {x - 1} \right)\\
e){x^2} - 6x - 16\\
= {x^2} - 8x + 2x - 16\\
= \left( {x - 8} \right)\left( {x + 2} \right)\\
f)3{x^2} - 16x + 5\\
= 3{x^2} - 15x - x + 5\\
= \left( {x - 5} \right)\left( {3x - 1} \right)\\
g)2{x^2} - 5x - 12\\
= 2{x^2} - 8x + 3x - 12\\
= \left( {x - 4} \right)\left( {2x + 3} \right)\\
m)4{x^2} - 12{x^2} - 16\\
= - 8{x^2} - 16\\
= - 8\left( {{x^2} + 1} \right)\\
n)\left( {x + 2} \right)\left( {x + 3} \right)\left( {x + 4} \right)\left( {x + 5} \right) - 24\\
= \left( {x + 2} \right)\left( {x + 5} \right)\left( {x + 3} \right)\left( {x + 4} \right) - 24\\
= \left( {{x^2} + 7x + 10} \right)\left( {{x^2} + 7x + 12} \right) - 24\\
Dat:\left( {{x^2} + 7x + 10} \right) = t\\
\Leftrightarrow t.\left( {t + 2} \right) - 24\\
= {t^2} + 2t - 24\\
= \left( {t + 6} \right)\left( {t - 2} \right)\\
= \left( {{x^2} + 7x + 10 + 6} \right)\left( {{x^2} + 7x + 10 - 2} \right)\\
= \left( {{x^2} + 7x + 16} \right)\left( {{x^2} + 7x + 8} \right)\\
= \left( {{x^2} + 7x + 16} \right)\left( {x + 1} \right)\left( {x + 7} \right)\\
h)\left( {{x^2} + 6x + 5} \right)\left( {{x^2} + 10x + 21} \right) + 15\\
= \left( {x + 1} \right)\left( {x + 5} \right)\left( {x + 3} \right)\left( {x + 7} \right) + 15\\
= \left( {x + 1} \right)\left( {x + 7} \right)\left( {x + 3} \right)\left( {x + 5} \right) + 15\\
= \left( {{x^2} + 8x + 7} \right)\left( {{x^2} + 8x + 15} \right) + 15\\
Dat:\left( {{x^2} + 8x + 7} \right) = t\\
\Leftrightarrow t\left( {t + 8} \right) + 15\\
= {t^2} + 8t + 15\\
= \left( {t + 3} \right)\left( {t + 5} \right)\\
= \left( {{x^2} + 8x + 7 + 3} \right)\left( {{x^2} + 8x + 7 + 5} \right)\\
= \left( {{x^2} + 8x + 10} \right)\left( {{x^2} + 8x + 12} \right)\\
= \left( {{x^2} + 8x + 7} \right)\left( {x + 2} \right)\left( {x + 6} \right)
\end{array}$
