Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
b,\\
{x^3} - 7x + 6 = \left( {{x^3} - {x^2}} \right) + \left( {{x^2} - x} \right) - \left( {6x - 6} \right)\\
 = {x^2}\left( {x - 1} \right) + x\left( {x - 1} \right) - 6.\left( {x - 1} \right)\\
 = \left( {x - 1} \right)\left( {{x^2} + x - 6} \right)\\
 = \left( {x - 1} \right).\left[ {\left( {{x^2} + 3x} \right) - \left( {2x + 6} \right)} \right]\\
 = \left( {x - 1} \right).\left[ {x.\left( {x + 3} \right) - 2.\left( {x + 3} \right)} \right]\\
 = \left( {x - 1} \right)\left( {x + 3} \right)\left( {x - 2} \right)\\
c,\\
{x^3} + 5{x^2} + 8x + 4\\
 = \left( {{x^3} + 4{x^2} + 4x} \right) + \left( {{x^2} + 4x + 4} \right)\\
 = x.\left( {{x^2} + 4x + 4} \right) + \left( {{x^2} + 4x + 4} \right)\\
 = \left( {{x^2} + 4x + 4} \right).\left( {x + 1} \right)\\
 = {\left( {x + 2} \right)^2}.\left( {x + 1} \right)\\
d,\\
{x^3} - 9{x^2} + 6x + 16\\
 = \left( {{x^3} - 8{x^2}} \right) - \left( {{x^2} - 8x} \right) - \left( {2x - 16} \right)\\
 = {x^2}\left( {x - 8} \right) - x.\left( {x - 8} \right) - 2.\left( {x - 8} \right)\\
 = \left( {x - 8} \right).\left( {{x^2} - x - 2} \right)\\
 = \left( {x - 8} \right).\left[ {\left( {{x^2} - 2x} \right) + \left( {x - 2} \right)} \right]\\
 = \left( {x - 8} \right).\left[ {x\left( {x - 2} \right) + \left( {x - 2} \right)} \right]\\
 = \left( {x - 8} \right)\left( {x - 2} \right)\left( {x + 1} \right)\\
e,\\
{x^3} - {x^2} - x - 2\\
 = \left( {{x^3} - 2{x^2}} \right) + \left( {{x^2} - 2x} \right) + \left( {x - 2} \right)\\
 = {x^2}\left( {x - 2} \right) + x\left( {x - 2} \right) + \left( {x - 2} \right)\\
 = \left( {x - 2} \right)\left( {{x^2} + x + 1} \right)\\
f,\\
{x^3} + {x^2} - x + 2\\
 = \left( {{x^3} + 2{x^2}} \right) - \left( {{x^2} + 2x} \right) + \left( {x + 2} \right)\\
 = {x^2}\left( {x + 2} \right) - x\left( {x + 2} \right) + \left( {x + 2} \right)\\
 = \left( {x + 2} \right)\left( {{x^2} - x + 1} \right)\\
h,\\
{x^3} - 6{x^2} - x + 30\\
 = \left( {{x^3} - 5{x^2}} \right) - \left( {{x^2} - 5x} \right) - \left( {6x - 30} \right)\\
 = {x^2}\left( {x - 5} \right) - x\left( {x - 5} \right) - 6.\left( {x - 5} \right)\\
 = \left( {x - 5} \right).\left( {{x^2} - x - 6} \right)\\
 = \left( {x - 5} \right).\left[ {\left( {{x^2} - 3x} \right) + \left( {2x - 6} \right)} \right]\\
 = \left( {x - 5} \right).\left[ {x\left( {x - 3} \right) + 2.\left( {x - 3} \right)} \right]\\
 = \left( {x - 5} \right)\left( {x - 3} \right)\left( {x + 2} \right)
\end{array}\)