Đáp án:
$\begin{array}{l}
8)4{x^2}y - 7y + 3xy\\
= y\left( {4{x^2} - 7 + 3x} \right)\\
= y.\left( {4{x^2} - 4x + 7x - 7} \right)\\
= y.\left( {x - 1} \right)\left( {4x + 7} \right)\\
9){x^2} - 5\\
= \left( {x - \sqrt 5 } \right)\left( {x + \sqrt 5 } \right)\\
10)\\
{x^3} - 4{x^2} - 12x + 27\\
= {x^3} + 27 - 4{x^2} - 12x\\
= \left( {x + 3} \right)\left( {{x^2} - 3x + 9} \right) - 4x\left( {x + 3} \right)\\
= \left( {x + 3} \right)\left( {{x^2} - 3x + 9 - 4x} \right)\\
= \left( {x + 3} \right)\left( {{x^2} - 7x + 9} \right)\\
11)\\
2{x^2} + 4x - 16\\
= 2.\left( {{x^2} + 2x - 8} \right)\\
= 2.\left( {{x^2} + 4x - 2x - 8} \right)\\
= 2.\left( {x + 4} \right)\left( {x - 2} \right)\\
12)\\
{x^2} + 3x + 2\\
= {x^2} + 2x + x + 2\\
= \left( {x + 2} \right)\left( {x + 1} \right)\\
13){x^2} + x - 6\\
= {x^2} + 2x - 3x - 6\\
= \left( {x + 2} \right)\left( {x - 3} \right)
\end{array}$
