Giải thích các bước giải:

$\begin{array}{l}
a){x^4} + 6{x^3} - 11{x^2} + 6x + 1\\
 = \left( {{x^4} + 1} \right) + \left( {6{x^3} + 6x} \right) - 11{x^2}\\
 = {\left( {{x^2} + 1} \right)^2} - 2{x^2} + 6x\left( {{x^2} + 1} \right) - 11{x^2}\\
 = {\left( {{x^2} + 1} \right)^2} + 6x\left( {{x^2} + 1} \right) - 13{x^2}\\
 = {\left( {{x^2} + 1} \right)^2} + 2\left( {{x^2} + 1} \right)3x + 9{x^2} - 22{x^2}\\
 = {\left( {{x^2} + 1 + 3x} \right)^2} - 22{x^2}\\
 = {\left( {{x^2} + 3x + 1} \right)^2} - {\left( {x\sqrt {22} } \right)^2}\\
 = \left( {{x^2} + 3x + 1 - x\sqrt {22} } \right)\left( {{x^2} + 3x + 1 + x\sqrt {22} } \right)\\
 = \left( {{x^2} + \left( {3 - \sqrt {22} } \right)x + 1} \right)\left( {{x^2} + \left( {3 + \sqrt {22} } \right)x + 1} \right)
\end{array}$

$\begin{array}{l}
b)6{a^4} + 7{a^3} - 37{a^2} - 8a + 12\\
 = \left( {6{a^4} - 12{a^3}} \right) + \left( {19{a^3} - 38{a^2}} \right) + \left( {{a^2} - 2a} \right) - \left( {6a - 12} \right)\\
 = 6{a^3}\left( {a - 2} \right) + 19{a^2}\left( {a - 2} \right) + a\left( {a - 2} \right) - 6\left( {a - 2} \right)\\
 = \left( {a - 2} \right)\left( {6{a^3} + 19{a^2} + a - 6} \right)\\
 = \left( {a - 2} \right)\left( {6{a^3} + 18{a^2} + {a^2} + 3a - 2a - 6} \right)\\
 = \left( {a - 2} \right)\left( {a + 3} \right)\left( {6{a^2} + a - 2} \right)\\
 = \left( {a - 2} \right)\left( {a + 3} \right)\left( {6{a^2} + 4a - 3a - 2} \right)\\
 = \left( {a - 2} \right)\left( {a + 3} \right)\left( {3a + 2} \right)\left( {2a - 1} \right)
\end{array}$