Đáp án:

\(\begin{array}{l}
a,\\
\left( {{x^2} - 2x + 2} \right)\left( {{x^2} + 2x + 2} \right)\\
\left( {x + 1} \right)\left( {x + 6} \right).\left( {{x^2} + 7x + 16} \right)\\
b,\\
\left[ \begin{array}{l}
x = 5\\
x =  - 6
\end{array} \right.
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a,\\
{x^4} + 4 = \left( {{x^4} + 4{x^2} + 4} \right) - 4{x^2}\\
 = \left[ {{{\left( {{x^2}} \right)}^2} + 2.{x^2}.2 + {2^2}} \right] - {\left( {2x} \right)^2}\\
 = {\left( {{x^2} + 2} \right)^2} - {\left( {2x} \right)^2}\\
 = \left[ {\left( {{x^2} + 2} \right) - 2x} \right].\left[ {\left( {{x^2} + 2} \right) + 2x} \right]\\
 = \left( {{x^2} - 2x + 2} \right)\left( {{x^2} + 2x + 2} \right)\\
\left( {x + 2} \right)\left( {x + 3} \right)\left( {x + 4} \right)\left( {x + 5} \right) - 24\\
 = \left[ {\left( {x + 2} \right)\left( {x + 5} \right)} \right].\left[ {\left( {x + 3} \right)\left( {x + 4} \right)} \right] - 24\\
 = \left( {{x^2} + 5x + 2x + 10} \right).\left( {{x^2} + 4x + 3x + 12} \right) - 24\\
 = \left( {{x^2} + 7x + 10} \right)\left( {{x^2} + 7x + 12} \right) - 24\\
 = \left[ {\left( {{x^2} + 7x + 11} \right) - 1} \right].\left[ {\left( {{x^2} + 7x + 11} \right) + 1} \right] - 24\\
 = {\left( {{x^2} + 7x + 11} \right)^2} - {1^2} - 24\\
 = {\left( {{x^2} + 7x + 11} \right)^2} - 25\\
 = {\left( {{x^2} + 7x + 11} \right)^2} - {5^2}\\
 = \left[ {\left( {{x^2} + 7x + 11} \right) - 5} \right].\left[ {\left( {{x^2} + 7x + 11} \right) + 5} \right]\\
 = \left( {{x^2} + 7x + 6} \right).\left( {{x^2} + 7x + 16} \right)\\
 = \left[ {\left( {{x^2} + x} \right) + \left( {6x + 6} \right)} \right].\left( {{x^2} + 7x + 16} \right)\\
 = \left[ {x\left( {x + 1} \right) + 6.\left( {x + 1} \right)} \right].\left( {{x^2} + 7x + 16} \right)\\
 = \left( {x + 1} \right)\left( {x + 6} \right).\left( {{x^2} + 7x + 16} \right)\\
b,\\
{x^4} - 30{x^2} + 31x - 30 = 0\\
 \Leftrightarrow \left( {{x^4} - 5{x^3}} \right) + \left( {5{x^3} - 25{x^2}} \right) + \left( { - 5{x^2} + 25x} \right) + \left( {6x - 30} \right) = 0\\
 \Leftrightarrow {x^3}\left( {x - 5} \right) + 5{x^2}\left( {x - 5} \right) - 5x\left( {x - 5} \right) + 6\left( {x - 5} \right) = 0\\
 \Leftrightarrow \left( {x - 5} \right).\left( {{x^3} + 5{x^2} - 5x + 6} \right) = 0\\
 \Leftrightarrow \left( {x - 5} \right).\left[ {\left( {{x^3} + 6{x^2}} \right) + \left( { - {x^2} - 6x} \right) + \left( {x + 6} \right)} \right] = 0\\
 \Leftrightarrow \left( {x - 5} \right).\left[ {{x^2}\left( {x + 6} \right) - x\left( {x + 6} \right) + \left( {x + 6} \right)} \right] = 0\\
 \Leftrightarrow \left( {x - 5} \right)\left( {x + 6} \right)\left( {{x^2} - x + 1} \right) = 0\\
{x^2} - x + 1 = \left( {{x^2} - x + \dfrac{1}{4}} \right) + \dfrac{3}{4} = \left( {{x^2} - 2.x.\dfrac{1}{2} + {{\left( {\dfrac{1}{2}} \right)}^2}} \right) + \dfrac{3}{4}\\
 = {\left( {x - \dfrac{1}{2}} \right)^2} + \dfrac{3}{4} \ge \dfrac{3}{4} > 0,\,\,\,\,\forall x\\
 \Rightarrow \left( {x - 5} \right)\left( {x + 6} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 5 = 0\\
x + 6 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 5\\
x =  - 6
\end{array} \right.
\end{array}\)