#### Answer

$(x^{n}+9)(x^{n}+11)$

#### Work Step by Step

For the moment, substitute $x^{n}=y.$
The trinomial becomes $y^{2}+20y+99.$
We search for integer factors of 99 whose sum is 20...
... and we find 9 and 11,
$=(y+9)(y+11)$
... back-substitute $x^{n}=y$
= $(x^{n}+9)(x^{n}+11)$