## Algebra 1: Common Core (15th Edition)

$2\sqrt3+3\sqrt 2$
We can simplify the product as follows: = $\sqrt 6(\sqrt 2+\sqrt 3)$ = $(\sqrt 2*\sqrt3)(\sqrt 2+\sqrt 3)$ = $(\sqrt 2*\sqrt3)*\sqrt 2+(\sqrt 2*\sqrt3)*\sqrt 3$ = $2\sqrt3+3\sqrt 2$ (since $\sqrt x * \sqrt y = \sqrt xy$)