Đáp án:

\(\begin{array}{l}
1,\\
\left( {x - 1} \right)\left( {x + 3} \right)\\
2,\\
\left( {x + 3} \right).\left( {3x - 2} \right)\\
3,\\
3.\left( {x - 1} \right)\left( {x + 2} \right)\\
4,\\
\left( {2x - 3} \right)\left( {3x - 2} \right)\\
5,\\
3.\left( {x + 2} \right).\left( {2x + 1} \right)\\
6,\\
2.\left( {x + 3} \right).\left( {3x + 1} \right)\\
7,\\
\left( {2x - 1} \right)\left( {4x + 3} \right)\\
8,\\
\left( {2x - 3} \right)\left( {4x + 1} \right)\\
9,\\
\left( { - x + 3} \right)\left( {8x + 1} \right)\\
10,\\
\left( {2x - 3} \right).\left( {5x + 2} \right)\\
11,\\
 - 2.\left( {x - 1} \right).\left( {5x + 3} \right)\\
12,\\
\left( { - 2x + 1} \right)\left( {5x + 6} \right)\\
13,\\
\left( {x + 2} \right).\left( { - 10x + 3} \right)\\
14,\\
2.\left( {x - 3} \right)\left( {5x + 1} \right)\\
15,\\
\left( { - 2x + 3} \right).\left( {5x + 4} \right)\\
16,\\
2.\left( {x - 1} \right)\left( {5x + 3} \right)
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
1,\\
{x^2} + 2x - 3\\
 = \left( {{x^2} - x} \right) + \left( {3x - 3} \right)\\
 = x.\left( {x - 1} \right) + 3.\left( {x - 1} \right)\\
 = \left( {x - 1} \right)\left( {x + 3} \right)\\
2,\\
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)\\
3,\\
3{x^2} + 3x - 6\\
 = 3.\left( {{x^2} + x - 2} \right)\\
 = 3.\left[ {\left( {{x^2} - x} \right) + \left( {2x - 2} \right)} \right]\\
 = 3.\left[ {x.\left( {x - 1} \right) + 2.\left( {x - 1} \right)} \right]\\
 = 3.\left( {x - 1} \right)\left( {x + 2} \right)\\
4,\\
6{x^2} - 13x + 6\\
 = \left( {6{x^2} - 9x} \right) + \left( { - 4x + 6} \right)\\
 = 3x\left( {2x - 3} \right) - 2.\left( {2x - 3} \right)\\
 = \left( {2x - 3} \right)\left( {3x - 2} \right)\\
5,\\
6{x^2} + 15x + 6\\
 = 3.\left( {2{x^2} + 5x + 2} \right)\\
 = 3.\left[ {\left( {2{x^2} + 4x} \right) + \left( {x + 2} \right)} \right]\\
 = 3.\left[ {2x.\left( {x + 2} \right) + \left( {x + 2} \right)} \right]\\
 = 3.\left( {x + 2} \right).\left( {2x + 1} \right)\\
6,\\
6{x^2} + 20x + 6\\
 = 2.\left( {3{x^2} + 10x + 3} \right)\\
 = 2.\left[ {\left( {3{x^2} + 9x} \right) + \left( {x + 3} \right)} \right]\\
 = 2.\left[ {3x.\left( {x + 3} \right) + \left( {x + 3} \right)} \right]\\
 = 2.\left( {x + 3} \right).\left( {3x + 1} \right)\\
7,\\
8{x^2} + 2x - 3\\
 = \left( {8{x^2} - 4x} \right) + \left( {6x - 3} \right)\\
 = 4x.\left( {2x - 1} \right) + 3.\left( {2x - 1} \right)\\
 = \left( {2x - 1} \right)\left( {4x + 3} \right)\\
8,\\
8{x^2} - 10x - 3\\
 = \left( {8{x^2} - 12x} \right) + \left( {2x - 3} \right)\\
 = 4x.\left( {2x - 3} \right) + \left( {2x - 3} \right)\\
 = \left( {2x - 3} \right)\left( {4x + 1} \right)\\
9,\\
 - 8{x^2} + 23x + 3\\
 = \left( { - 8{x^2} + 24x} \right) + \left( { - x + 3} \right)\\
 = 8x.\left( { - x + 3} \right) + \left( { - x + 3} \right)\\
 = \left( { - x + 3} \right)\left( {8x + 1} \right)\\
10,\\
10{x^2} - 11x - 6\\
 = \left( {10{x^2} - 15x} \right) + \left( {4x - 6} \right)\\
 = 5x.\left( {2x - 3} \right) + 2.\left( {2x - 3} \right)\\
 = \left( {2x - 3} \right).\left( {5x + 2} \right)\\
11,\\
 - 10{x^2} + 4x + 6\\
 =  - 2.\left( {5{x^2} - 2x - 3} \right)\\
 =  - 2.\left[ {\left( {5{x^2} - 5x} \right) + \left( {3x - 3} \right)} \right]\\
 =  - 2.\left[ {5x.\left( {x - 1} \right) + 3.\left( {x - 1} \right)} \right]\\
 =  - 2.\left( {x - 1} \right).\left( {5x + 3} \right)\\
12,\\
 - 10{x^2} - 7x + 6\\
 = \left( { - 10{x^2} + 5x} \right) + \left( { - 12x + 6} \right)\\
 = 5x.\left( { - 2x + 1} \right) + 6.\left( { - 2x + 1} \right)\\
 = \left( { - 2x + 1} \right)\left( {5x + 6} \right)\\
13,\\
 - 10{x^2} - 17x + 6\\
 = \left( { - 10{x^2} - 20x} \right) + \left( {3x + 6} \right)\\
 =  - 10x.\left( {x + 2} \right) + 3.\left( {x + 2} \right)\\
 = \left( {x + 2} \right).\left( { - 10x + 3} \right)\\
14,\\
10{x^2} - 28x - 6\\
 = 2.\left( {5{x^2} - 14x - 3} \right)\\
 = 2.\left[ {\left( {5{x^2} - 15x} \right) + \left( {x - 3} \right)} \right]\\
 = 2.\left[ {5x.\left( {x - 3} \right) + \left( {x - 3} \right)} \right]\\
 = 2.\left( {x - 3} \right)\left( {5x + 1} \right)\\
15,\\
 - 10{x^2} + 7x + 12\\
 = \left( { - 10{x^2} + 15x} \right) + \left( { - 8x + 12} \right)\\
 = 5x.\left( { - 2x + 3} \right) + 4.\left( { - 2x + 3} \right)\\
 = \left( { - 2x + 3} \right).\left( {5x + 4} \right)\\
16,\\
10{x^2} - 4x - 6\\
 = 2.\left( {5{x^2} - 2x - 3} \right)\\
 = 2.\left[ {\left( {5{x^2} - 5x} \right) + \left( {3x - 3} \right)} \right]\\
 = 2.\left[ {5x.\left( {x - 1} \right) + 3.\left( {x - 1} \right)} \right]\\
 = 2.\left( {x - 1} \right)\left( {5x + 3} \right)
\end{array}\)