#### Answer

$(x^{2}+1)^{5}(x+3)^{3}(x^{2}+x+4)$

#### Work Step by Step

We can factor out common factors:
$(x^{2}+1)^{5}$ and $(x+3)^{3}$
$..=(x^{2}+1)^{5}(x+3)^{3}[(x+3)+(x^{2}+1)]$
$=(x^{2}+1)^{5}(x+3)^{3}(x^{2}+x+4)$
$x^{2}+1$ can not be further factored,
as there is no special formula for a sum of two squares.
For the last trinomial, we search for integer factors of 4, that add to 1... none,
so we leave it as a prime (irreducible) factor.