Giải thích các bước giải:
$\begin{array}{l}
15)4{x^4} - 21{x^2}{y^2} + {y^4}\\
= \left( {4{x^4} - 4{x^2}{y^2} + {y^4}} \right) - 17{x^2}{y^2}\\
= {\left( {2{x^2} - {y^2}} \right)^2} - {\left( {xy\sqrt {17} } \right)^2}\\
= \left( {2{x^2} - xy\sqrt {17} - {y^2}} \right)\left( {2{x^2} + xy\sqrt {17} - {y^2}} \right)\\
16)81{x^4} + 4\\
= {\left( {9{x^2}} \right)^2} + 2.9{x^2}.2 + 4 - 36{x^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)\\
17){x^5} + {x^4} + 1\\
= \left( {{x^5} - {x^2}} \right) + \left( {{x^4} + {x^2} + 1} \right)\\
= {x^2}\left( {{x^3} - 1} \right) + \left( {{x^4} + 2{x^2} + 1 - {x^2}} \right)\\
= {x^2}\left( {x - 1} \right)\left( {{x^2} + x + 1} \right) + \left( {{{\left( {{x^2} + 1} \right)}^2} - {x^2}} \right)\\
= {x^2}\left( {x - 1} \right)\left( {{x^2} + x + 1} \right) + \left( {{x^2} - x + 1} \right)\left( {{x^2} + x + 1} \right)\\
= \left( {{x^2} + x + 1} \right)\left( {{x^2}\left( {x - 1} \right) + {x^2} - x + 1} \right)\\
18){\left( {{x^2} + x} \right)^2} - 2\left( {{x^2} + x} \right) - 63\\
= {\left( {{x^2} + x - 1} \right)^2} - 64\\
= \left( {{x^2} + x - 9} \right)\left( {{x^2} + x + 7} \right)\\
19)\left( {x + 2} \right)\left( {x + 3} \right)\left( {x + 4} \right)\left( {x + 5} \right) - 24\\
= \left( {{x^2} + 7x + 10} \right)\left( {{x^2} + 7x + 12} \right) - 24\\
= {\left( {{x^2} + 7x + 10} \right)^2} + 2\left( {{x^2} + 7x + 10} \right) - 24\\
= \left( {{x^2} + 7x + 11 - 5} \right)\left( {{x^2} + 7x + 11 + 5} \right)\\
= \left( {{x^2} + 7x + 6} \right)\left( {{x^2} + 7x + 16} \right)\\
= \left( {x + 1} \right)\left( {x + 6} \right)\left( {{x^2} + 7x + 16} \right)\\
20)\left( {{x^2} + 8x + 7} \right)\left( {{x^2} + 8x + 15} \right) + 15\\
= {\left( {{x^2} + 8x + 7} \right)^2} + 8\left( {{x^2} + 8x + 7} \right) + 15\\
= {\left( {{x^2} + 8x + 13} \right)^2} - 1\\
= \left( {{x^2} + 8x + 12} \right)\left( {{x^2} + 8x + 14} \right)\\
21){x^3} + 4{x^2} + 5x + 2\\
= {x^2}\left( {x + 1} \right) + 3x\left( {x + 1} \right) + 2\left( {x + 1} \right)\\
= \left( {x + 1} \right)\left( {{x^2} + 3x + 2} \right)\\
= {\left( {x + 1} \right)^2}\left( {x + 2} \right)\\
22)2{x^4} - 3{x^3} - 7{x^2} + 6x + 8\\
= 2{x^3}\left( {x + 1} \right) - 5{x^2}\left( {x + 1} \right) - 2x\left( {x + 1} \right) + 8\left( {x + 1} \right)\\
= \left( {x + 1} \right)\left( {2{x^3} - 5{x^2} - 2x + 8} \right)\\
= \left( {x + 1} \right)\left( {2{x^2}\left( {x - 2} \right) - x\left( {x - 2} \right) - 4\left( {x - 2} \right)} \right)\\
= \left( {x + 1} \right)\left( {x - 2} \right)\left( {2{x^2} - x - 4} \right)\\
23){\left( {{a^2} + {b^2} - 5} \right)^2} - 4{\left( {ab + 2} \right)^2}\\
= {\left( {{a^2} + {b^2} - 5} \right)^2} - {\left( {2ab + 4} \right)^2}\\
= \left( {{a^2} + {b^2} - 5 - 2ab - 4} \right)\left( {{a^2} + {b^2} - 5 + 2ab + 4} \right)\\
= \left( {{{\left( {a - b} \right)}^2} - 9} \right)\left( {{{\left( {a + b} \right)}^2} - 1} \right)\\
= \left( {a - b - 3} \right)\left( {a - b + 3} \right)\left( {a + b - 1} \right)\left( {a + b + 1} \right)
\end{array}$
