Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
a,\\
{a^3} - 7a - 6\\
 = \left( {{a^3} + {a^2}} \right) - \left( {{a^2} + a} \right) - \left( {6a + 6} \right)\\
 = {a^2}\left( {a + 1} \right) - a\left( {a + 1} \right) - 6\left( {a + 1} \right)\\
 = \left( {a + 1} \right)\left( {{a^2} - a - 6} \right)\\
 = \left( {a + 1} \right).\left[ {\left( {{a^2} - 3a} \right) + \left( {2a - 6} \right)} \right]\\
 = \left( {a + 1} \right).\left[ {a\left( {a - 3} \right) + 2\left( {a - 3} \right)} \right]\\
 = \left( {a + 1} \right)\left( {a - 3} \right)\left( {a + 2} \right)\\
b,\\
{a^3} + 4{a^2} - 7a - 10\\
 = \left( {{a^3} + {a^2}} \right) + \left( {3{a^2} + 3a} \right) - \left( {10a + 10} \right)\\
 = {a^2}\left( {a + 1} \right) + 3a\left( {a + 1} \right) - 10\left( {a + 1} \right)\\
 = \left( {a + 1} \right)\left( {{a^2} + 3a - 10} \right)\\
 = \left( {a + 1} \right)\left[ {\left( {{a^2} + 5a} \right) - \left( {2a + 10} \right)} \right]\\
 = \left( {a + 1} \right)\left[ {a\left( {a + 5} \right) - 2\left( {a + 5} \right)} \right]\\
 = \left( {a + 1} \right)\left( {a + 5} \right)\left( {a - 2} \right)\\
d,\\
{\left( {{a^2} + a} \right)^2} + 4.\left( {{a^2} + a} \right) - 12\\
 = \left[ {{{\left( {{a^2} + a} \right)}^2} + 6\left( {{a^2} + a} \right)} \right] - \left[ {2\left( {{a^2} + a} \right) + 12} \right]\\
 = \left( {{a^2} + a} \right).\left[ {\left( {{a^2} + a} \right) + 6} \right] - 2.\left[ {\left( {{a^2} + a} \right) + 6} \right]\\
 = \left( {{a^2} + a + 6} \right)\left[ {\left( {{a^2} + a} \right) - 2} \right]\\
 = \left( {{a^2} + a + 6} \right).\left[ {\left( {{a^2} - a} \right) + \left( {2a - 2} \right)} \right]\\
 = \left( {{a^2} + a + 6} \right).\left[ {a\left( {a - 1} \right) + 2\left( {a - 1} \right)} \right]\\
 = \left( {{a^2} + a + 6} \right).\left( {a - 1} \right)\left( {a + 2} \right)\\
e,\\
\left( {{x^2} + x + 1} \right)\left( {{x^2} + x + 2} \right) - 12\\
 = \left( {{x^2} + x + 1} \right).\left[ {\left( {{x^2} + x + 1} \right) + 1} \right] - 12\\
 = {\left( {{x^2} + x + 1} \right)^2} + \left( {{x^2} + x + 1} \right) - 12\\
 = \left[ {{{\left( {{x^2} + x + 1} \right)}^2} + 4.\left( {{x^2} + x + 1} \right)} \right] - \left[ {3.\left( {{x^2} + x + 1} \right) + 12} \right]\\
 = \left( {{x^2} + x + 1} \right).\left[ {\left( {{x^2} + x + 1} \right) + 4} \right] - 3.\left[ {\left( {{x^2} + x + 1} \right) + 4} \right]\\
 = \left( {{x^2} + x + 5} \right).\left[ {\left( {{x^2} + x + 1} \right) - 3} \right]\\
 = \left( {{x^2} + x + 5} \right).\left( {{x^2} + x - 2} \right)\\
 = \left( {{x^2} + x + 5} \right).\left[ {\left( {{x^2} - x} \right) + \left( {2x - 2} \right)} \right]\\
 = \left( {{x^2} + x + 5} \right).\left[ {x\left( {x - 1} \right) + 2\left( {x - 1} \right)} \right]\\
 = \left( {{x^2} + x + 5} \right).\left( {x - 1} \right)\left( {x + 2} \right)
\end{array}\) 

Em xem lại đề câu c nhé!