Answer
\[\left( {{x}^{n}}+9 \right)\left( {{x}^{n}}+11 \right)\].
Work Step by Step
Put \[{{x}^{n}}=p\]then the equation becomes
\[\begin{align}
& {{p}^{2}}+20p+99={{p}^{2}}+9p+11p+99 \\
& =p\left( p+9 \right)+11\left( p+9 \right) \\
& =\left( p+9 \right)\left( p+11 \right)
\end{align}\]
So,
\[\begin{align}
& p={{x}^{n}} \\
& \left( {{x}^{n}}+9 \right)\left( {{x}^{n}}+11 \right) \\
\end{align}\]
The given Polynomial can be factored as\[\left( {{x}^{n}}+9 \right)\left( {{x}^{n}}+11 \right)\].
