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

$105c\sqrt 3c$
Solving and simplifying the expression: $3\sqrt 5c \times7\sqrt 15c^{2}$ = $(3\times7)(\sqrt 5c \times\sqrt 15c^{2})$ = $21(\sqrt 15 \times \sqrt 5 \times \sqrt c \times \sqrt c^{2})$ = $21(\sqrt 75 \times\sqrt c^{3})$ = $21(\sqrt 75c^{3})$ Find the square factor of $(\sqrt 75c^{3})$ $\sqrt 75c^{3} = 3 \times25 \times c^2 \times c$ =$3 \times5^2 \times c^2 \times c$ = $3c \times5^2 \times c^2$ =$3c (5 c)^2$ So $25c^2$ is its square factor Put it in the equation: = $21(\sqrt 3c\times\sqrt 5c^{2})$ = $21(5c \sqrt 3c )$ = $21\times 5c (\sqrt 3c)$ = $105c \sqrt 3c$