Finite Math and Applied Calculus (6th Edition)

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