phân tích đa thức thành nhân tử a. 4x^2 – 4x -3 b. x^2 – x – 6 c. 3x^2 – 16x + 5 d. 2x^2 – 5x -12 e. 6x^2 – 7x – 20

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
a,\\
4{x^2} – 4x – 3\\
 = \left( {4{x^2} – 6x} \right) + \left( {2x – 3} \right)\\
 = 2x.\left( {2x – 3} \right) + \left( {2x – 3} \right)\\
 = \left( {2x – 3} \right)\left( {2x + 1} \right)\\
b,\\
{x^2} – x – 6\\
 = \left( {{x^2} – 3x} \right) + \left( {2x – 6} \right)\\
 = x\left( {x – 3} \right) + 2\left( {x – 3} \right)\\
 = \left( {x – 3} \right)\left( {x + 2} \right)\\
c,\\
3{x^2} – 16x + 5\\
 = \left( {3{x^2} – 15x} \right) – \left( {x – 5} \right)\\
 = 3x\left( {x – 5} \right) – \left( {x – 5} \right)\\
 = \left( {x – 5} \right)\left( {3x – 1} \right)\\
d,\\
2{x^2} – 5x – 12\\
 = \left( {2{x^2} – 8x} \right) + \left( {3x – 12} \right)\\
 = 2x\left( {x – 4} \right) + 3\left( {x – 4} \right)\\
 = \left( {x – 4} \right)\left( {2x + 3} \right)\\
e,\\
6{x^2} – 7x – 20\\
 = \left( {6{x^2} – 15x} \right) + \left( {8x – 20} \right)\\
 = 3x.\left( {2x – 5} \right) + 4.\left( {2x – 5} \right)\\
 = \left( {2x – 5} \right)\left( {3x + 4} \right)
\end{array}\)