Đáp án:
$\begin{array}{l}
2)a){x^2} + 3x + 2x + 6\\
= \left( {x + 3} \right)\left( {x + 2} \right)\\
b){x^2} + 5x + 6\\
= \left( {x + 2} \right)\left( {x + 3} \right)\\
c){x^2} + 9x + 18\\
= \left( {x + 3} \right)\left( {x + 6} \right)\\
d){x^2} - x - 20\\
= \left( {x - 5} \right)\left( {x + 4} \right)\\
3)a){x^3} - 9x = 0\\
\Rightarrow x\left( {{x^2} - 9} \right) = 0\\
\Rightarrow x\left( {x - 3} \right)\left( {x + 3} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x = 0\\
x = 3\\
x = - 3
\end{array} \right.\\
Vậy\,x = 0;x = 3;x = - 3\\
b){x^3} + 4{x^2} + 4x = 0\\
\Rightarrow x\left( {{x^2} + 4x + 4} \right) = 0\\
\Rightarrow x{\left( {x + 2} \right)^2} = 0\\
\Rightarrow \left[ \begin{array}{l}
x = 0\\
x = - 2
\end{array} \right.\\
Vậy\,x = 0;x = - 2\\
4){x^3} - {x^2} - 25x + 25 = 0\\
\Rightarrow \left( {x - 1} \right)\left( {{x^2} - 25} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x = 1\\
x = 5\\
x = - 5
\end{array} \right.\\
b)4{x^3} - 8{x^2} - 9x + 18 = 0\\
\Rightarrow \left( {x - 2} \right)\left( {4{x^2} - 9} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x = 2\\
x = \pm \frac{3}{2}
\end{array} \right.\\
8)a)12{x^2} - 72x + 60\\
= 12\left( {{x^2} - 6x + 5} \right)\\
= 12\left( {x - 1} \right)\left( {x - 5} \right)\\
b)9{x^2} - 21x - 18\\
= 3\left( {3{x^2} - 7x - 6} \right)\\
= 3\left( {3{x^2} - 9x + 2x - 6} \right)\\
= 3\left( {3x + 2} \right)\left( {x - 3} \right)
\end{array}$
