## Introductory Algebra for College Students (7th Edition)

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