Đáp án:
$\begin{array}{l}
14)\\
4{x^2} + 4x - 3\\
= 4{x^2} + 4x + 1 - 4\\
= {\left( {2x + 1} \right)^2} - 4\\
= \left( {2x + 1 - 2} \right)\left( {2x + 1 + 2} \right)\\
= \left( {2x - 1} \right)\left( {2x + 3} \right)\\
15)4{x^2} + 15x + 9\\
= 4{x^2} + 12x + 3x + 9\\
= \left( {x + 3} \right)\left( {4x + 3} \right)\\
16)4{x^2} - 7x - 2\\
= 4{x^2} - 8x + x - 2\\
= \left( {x - 2} \right)\left( {4x + 1} \right)\\
17)3{x^2} + x - 2\\
= 3{x^2} + 3x - 2x - 2\\
= \left( {x + 1} \right)\left( {3x - 2} \right)\\
18)\\
6{x^2} + 7x + 2\\
= 6{x^2} + 3x + 4x + 2\\
= \left( {2x + 1} \right)\left( {3x + 2} \right)\\
19)5{x^2} - 18x - 8\\
= 5{x^2} - 20x + 2x - 8\\
= \left( {x - 4} \right)\left( {5x + 2} \right)\\
20)\\
{x^2} - xy - 2{y^2}\\
= {x^2} - 2xy + xy - 2{y^2}\\
= \left( {x - 2y} \right)\left( {x + y} \right)\\
21){x^2} - 3xy + 2{y^2}\\
= {x^2} - xy - 2xy + 2{y^2}\\
= \left( {x - y} \right)\left( {x - 2y} \right)\\
22)2{x^2} - 3xy - 2{y^2}\\
= 2{x^2} - 4xy + xy - 2{y^2}\\
= \left( {x - 2y} \right)\left( {2x + y} \right)\\
23)2{x^2} + 5xy + 2{y^2}\\
= 2{x^2} + 4xy + xy + 2{y^2}\\
= \left( {x + 2y} \right)\left( {2x + y} \right)\\
24)2{x^2} + 2xy - 4{y^2}\\
= 2\left( {{x^2} + xy - 2{y^2}} \right)\\
= 2.\left( {{x^2} + 2xy - xy - 2{y^2}} \right)\\
= 2.\left( {x + 2y} \right)\left( {x - y} \right)
\end{array}$
