Đáp án:
$\begin{array}{l}
a)2{x^8} = 2.{x^8}\\
b){x^2} - 8x - 9\\
= {x^2} - 9x + x - 9\\
= x\left( {x - 9} \right) + x - 9\\
= \left( {x - 9} \right)\left( {x + 1} \right)\\
c){x^2} + 14x + 48\\
= {x^2} + 6x + 8x + 48\\
= x\left( {x + 6} \right) + 8\left( {x + 6} \right)\\
= \left( {x + 6} \right)\left( {x + 8} \right)\\
d)4{x^4} - 21{x^2}{y^2} + {y^4}\\
= {\left( {2{x^2}} \right)^2} - 2.2{x^2}.\frac{{21}}{4}{y^2} + \frac{{441}}{{16}}{y^4} - \frac{{425}}{{16}}{y^4}\\
= {\left( {2{x^2} - \frac{{21}}{4}{y^2}} \right)^2} - \frac{{425}}{{16}}{y^4}\\
= \left( {2{x^2} - \frac{{21}}{4}{y^2} - \frac{{\sqrt {425} }}{4}{y^2}} \right)\left( {2{x^2} - \frac{{21}}{4}{y^2} + \frac{{\sqrt {425} }}{4}{y^2}} \right)\\
e){x^5} - 5{x^3} + 4x\\
= x\left( {{x^4} - 5{x^2} + 4} \right)\\
= x\left( {{x^4} - {x^2} - 4{x^2} + 4} \right)\\
= x\left( {{x^2} - 1} \right)\left( {{x^2} - 4} \right)\\
= x\left( {x - 1} \right)\left( {x + 1} \right)\left( {x - 2} \right)\left( {x + 2} \right)
\end{array}$
