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

$16y^3$
Simpifying $\sqrt 2y \times \sqrt 128y^5$ $\sqrt 2 \times \sqrt 128 \times\sqrt y \times \sqrt y^5$ = $\sqrt 256y^6$ Finding the perfect square factor of $256y^6$ $256y^6 = 4 \times 64 \times y^6$ = $2^2 \times8^2 \times y^6$ = $(2 \times8 \times y^3)^2$ = $(16y^3)^2$ $\sqrt 256y^6 = \sqrt (16y^3)^2$ =$\sqrt16y^3$