Giải thích các bước giải:

$\begin{array}{l}
1)5ax - 15ay + 20a = 5a\left( {x - 3y + 4} \right)\\
2)5a{x^2} - 3axy + 3a{y^2} - 5axy\\
 = a\left( {5{x^2} - 3xy + 3{y^2} - 5xy} \right)\\
 = a\left( {\left( {5{x^2} - 5xy} \right) - \left( {3xy - 3{y^2}} \right)} \right)\\
 = a\left( {5x\left( {x - y} \right) - 3y\left( {x - y} \right)} \right)\\
 = a\left( {x - y} \right)\left( {5x - 3y} \right)\\
3)10ax - 5ay - 2x + y\\
 = \left( {10ax - 5ay} \right) - \left( {2x - y} \right)\\
 = 5a\left( {2x - y} \right) - \left( {2x - y} \right)\\
 = \left( {2x - y} \right)\left( {5a - 1} \right)\\
4){\left( {x + 7} \right)^2} - {\left( {3x + 8} \right)^2}\\
 = \left( {x + 7 - \left( {3x + 8} \right)} \right)\left( {x + 7 + \left( {3x + 8} \right)} \right)\\
 = \left( { - 2x - 1} \right)\left( {4x + 15} \right)\\
 =  - \left( {2x + 1} \right)\left( {4x + 15} \right)\\
5)\frac{4}{9}{a^4} - \frac{{25}}{4}\\
 = {\left( {\frac{2}{3}{a^2}} \right)^2} - {\left( {\frac{5}{2}} \right)^2}\\
 = \left( {\frac{2}{3}{a^2} - \frac{5}{2}} \right)\left( {\frac{2}{3}{a^2} + \frac{5}{2}} \right)\\
6){x^2} - 8xy + 16{y^2} - 36{b^2}\\
 = {x^2} - 2x.4y + {\left( {4y} \right)^2} - {\left( {6b} \right)^2}\\
 = {\left( {x - 4y} \right)^2} - {\left( {6b} \right)^2}\\
 = \left( {x - 4y - 6b} \right)\left( {x - 4y + 6b} \right)\\
7){x^2} + 4xy + 4{y^2} - 9{x^2} + 6x - 1\\
 = \left( {{x^2} + 2.x.2y + {{\left( {2y} \right)}^2}} \right) - \left( {{{\left( {3x} \right)}^2} - 2.3x + 1} \right)\\
 = {\left( {x + 2y} \right)^2} - {\left( {3x - 1} \right)^2}\\
 = \left( {x + 2y - \left( {3x - 1} \right)} \right)\left( {x + 2y + 3x - 1} \right)\\
 = \left( {2y - 2x + 1} \right)\left( {4x + 2y - 1} \right)\\
8)49 - {a^2} + 10a - 25\\
 = {7^2} - \left( {{a^2} - 2.a.5 + {5^2}} \right)\\
 = {7^2} - {\left( {a - 5} \right)^2}\\
 = \left( {7 - \left( {a - 5} \right)} \right)\left( {7 + a - 5} \right)\\
 = \left( {12 - a} \right)\left( {a + 2} \right)\\
9){x^2} - x - 20\\
 = \left( {{x^2} - 5x} \right) + \left( {4x - 20} \right)\\
 = x\left( {x - 5} \right) + 4\left( {x - 5} \right)\\
 = \left( {x - 5} \right)\left( {x + 4} \right)\\
10)4{x^2} + 15x + 9\\
 = \left( {4{x^2} + 12x} \right) + \left( {3x + 9} \right)\\
 = 4x\left( {x + 3} \right) + 3\left( {x + 3} \right)\\
 = \left( {x + 3} \right)\left( {4x + 3} \right)\\
11)3{x^2} + 7x - 6\\
 = \left( {3{x^2} + 9x} \right) - \left( {2x + 6} \right)\\
 = 3x\left( {x + 3} \right) - 2\left( {x + 3} \right)\\
 = \left( {x + 3} \right)\left( {3x - 2} \right)\\
12)3{x^2} + 10x + 3\\
 = \left( {3{x^2} + 9x} \right) + \left( {x + 3} \right)\\
 = 3x\left( {x + 3} \right) + \left( {x + 3} \right)\\
 = \left( {x + 3} \right)\left( {3x + 1} \right)
\end{array}$