Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
{x^2} - 6x - 6\\
= \left( {{x^2} - 6x + 9} \right) - 15\\
= {\left( {x - 3} \right)^2} - 15\\
= \left( {x - 3 - \sqrt {15} } \right)\left( {x - 3 + \sqrt {15} } \right)\\
b,\\
{x^4} + 4{x^2} - 5\\
= \left( {{x^4} - {x^2}} \right) + \left( {5{x^2} - 5} \right)\\
= {x^2}\left( {{x^2} - 1} \right) + 5.\left( {{x^2} - 1} \right)\\
= \left( {{x^2} - 1} \right)\left( {{x^2} + 5} \right)\\
= \left( {x - 1} \right)\left( {x + 1} \right)\left( {{x^2} + 5} \right)\\
c,\\
{x^3} - 19x - 30\\
= \left( {{x^3} - 5{x^2}} \right) + \left( {5{x^2} - 25x} \right) + \left( {6x - 30} \right)\\
= {x^2}\left( {x - 5} \right) + 5x\left( {x - 5} \right) + 6.\left( {x - 5} \right)\\
= \left( {x - 5} \right)\left( {{x^2} + 5x + 6} \right)\\
= \left( {x - 5} \right).\left[ {\left( {{x^2} + 2x} \right) + \left( {3x + 6} \right)} \right]\\
= \left( {x - 5} \right).\left[ {x\left( {x + 2} \right) + 3.\left( {x + 2} \right)} \right]\\
= \left( {x - 5} \right)\left( {x + 2} \right)\left( {x + 3} \right)\\
d,\\
{\left( {1 + {x^2}} \right)^2} - 4x.\left( {1 - {x^2}} \right)\\
= \left( {1 + 2{x^2} + {x^4}} \right) - \left( {4x - 4{x^3}} \right)\\
= {x^4} + 2{x^2} + 1 - 4x + 4{x^3}\\
= {x^4} + 4{x^3} + 2{x^2} - 4x + 1\\
= \left( {{x^4} + 2{x^3} - {x^2}} \right) + \left( {2{x^3} + 4{x^2} - 2x} \right) + \left( { - {x^2} - 2x + 1} \right)\\
= {x^2}\left( {{x^2} + 2x - 1} \right) + 2x.\left( {{x^2} + 2x - 1} \right) - \left( {{x^2} - 2x + 1} \right)\\
= \left( {{x^2} + 2x - 1} \right).\left( {{x^2} + 2x - 1} \right)\\
= {\left( {{x^2} - 2x + 1} \right)^2}\\
e,\\
{\left( {{x^2} - 8} \right)^2} + 36\\
= {x^4} - 2.{x^2}.8 + {8^2} + 36\\
= {x^4} - 16{x^2} + 100\\
= \left( {{x^4} + 20{x^2} + 100} \right) - 36{x^2}\\
= {\left( {{x^2} + 10} \right)^2} - {\left( {6x} \right)^2}\\
= \left( {{x^2} - 6x + 10} \right)\left( {{x^2} + 6x + 10} \right)\\
g,\\
81{x^4} + 4\\
= \left( {81{x^4} + 36{x^2} + 4} \right) - 36{x^2}\\
= \left[ {{{\left( {9{x^2}} \right)}^2} + 2.9{x^2}.2 + {2^2}} \right] - {\left( {6x} \right)^2}\\
= {\left( {9{x^2} + 2} \right)^2} - {\left( {6x} \right)^2}\\
= \left( {9{x^2} - 6x + 2} \right)\left( {9{x^2} + 6x + 2} \right)\\
h,\\
{x^5} + x + 1\\
= \left( {{x^5} + {x^4} + {x^3}} \right) - \left( {{x^4} + {x^3} + {x^2}} \right) + \left( {{x^2} + x + 1} \right)\\
= {x^3}\left( {{x^2} + x + 1} \right) - {x^2}\left( {{x^2} + x + 1} \right) + \left( {{x^2} + x + 1} \right)\\
= \left( {{x^2} + x + 1} \right)\left( {{x^3} - {x^2} + 1} \right)
\end{array}\)
