Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
4{x^2} - 4x - 3\\
= \left( {4{x^2} - 6x} \right) + \left( {2x - 3} \right)\\
= 2x.\left( {2x - 3} \right) + \left( {2x - 3} \right)\\
= \left( {2x - 3} \right)\left( {2x + 1} \right)\\
b,\\
{x^2} - x - 6\\
= \left( {{x^2} - 3x} \right) + \left( {2x - 6} \right)\\
= x\left( {x - 3} \right) + 2\left( {x - 3} \right)\\
= \left( {x - 3} \right)\left( {x + 2} \right)\\
c,\\
3{x^2} - 16x + 5\\
= \left( {3{x^2} - 15x} \right) - \left( {x - 5} \right)\\
= 3x\left( {x - 5} \right) - \left( {x - 5} \right)\\
= \left( {x - 5} \right)\left( {3x - 1} \right)\\
d,\\
2{x^2} - 5x - 12\\
= \left( {2{x^2} - 8x} \right) + \left( {3x - 12} \right)\\
= 2x\left( {x - 4} \right) + 3\left( {x - 4} \right)\\
= \left( {x - 4} \right)\left( {2x + 3} \right)\\
e,\\
6{x^2} - 7x - 20\\
= \left( {6{x^2} - 15x} \right) + \left( {8x - 20} \right)\\
= 3x.\left( {2x - 5} \right) + 4.\left( {2x - 5} \right)\\
= \left( {2x - 5} \right)\left( {3x + 4} \right)
\end{array}\)
