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

$(x^{3}-5)(1+2y)$
There are no common factors (for all four terms). Reorder and factor by grouping: $...=x^{3}+2x^{3}y-5-10y$ $=x^{3}\cdot 1+x^{3}\cdot 2y+(-5)\cdot 1+(-5)\cdot 2y$ ... use distribution, group by group $=x^{3}(1+2y)+(-5)(1+2y)$ ... use distribution, common factor: $(1+2y)$ = $(x^{3}-5)(1+2y)$