Đáp án:
$\begin{array}{l}
1)a)15{x^2} + 7x - 2\\
= 15{x^2} + 10x - 3x - 2\\
= 5x\left( {3x + 2} \right) - \left( {3x + 2} \right)\\
= \left( {3x + 2} \right)\left( {5x - 1} \right)\\
b)2{x^2} + 5x + 2\\
= 2{x^2} + x + 4x + 2\\
= x\left( {2x + 1} \right) + 2\left( {2x + 1} \right)\\
= \left( {2x + 1} \right)\left( {x + 2} \right)\\
B2)\\
a){x^4} + {x^2} - 2\\
= {x^4} + 2{x^2} - {x^2} - 2\\
= \left( {{x^2} + 2} \right)\left( {{x^2} - 1} \right)\\
= \left( {{x^2} + 2} \right)\left( {x - 1} \right)\left( {x + 1} \right)\\
b){x^4} + 4{x^2} - 5\\
= {x^4} + 5{x^2} - {x^2} - 5\\
= \left( {{x^2} + 5} \right)\left( {{x^2} - 1} \right)\\
= \left( {{x^2} + 5} \right)\left( {x + 1} \right)\left( {x - 1} \right)\\
c){x^3} - 19x - 30\\
= {x^3} - 5{x^2} + 5{x^2} - 25x + 6x - 30\\
= \left( {x - 5} \right)\left( {{x^2} + 5x + 6} \right)\\
= \left( {x - 5} \right)\left( {{x^2} + 2x + 3x + 6} \right)\\
= \left( {x - 5} \right)\left( {x + 2} \right)\left( {x + 3} \right)\\
d){x^3} - 7x - 6\\
= {x^3} + {x^2} - {x^2} - x - 6x - 6\\
= \left( {x + 1} \right)\left( {{x^2} - x - 6} \right)\\
= \left( {x + 1} \right)\left( {x - 3} \right)\left( {x + 2} \right)\\
e){x^3} - 5{x^2} - 14x\\
= x\left( {{x^2} - 5x - 14} \right)\\
= x\left( {{x^2} - 7x + 2x - 14} \right)\\
= x\left( {x - 7} \right)\left( {x + 2} \right)
\end{array}$
