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

Published by Prentice Hall

# Chapter 10 - Radical Expressions and Equations - 10-2 Simplifying Radicals - Practice and Problem-Solving Exercises - Page 623: 28

#### Answer

$105c\sqrt 3c$

#### Work Step by Step

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: = $21(\sqrt 3c\times\sqrt 5c^{2})$ = $21(5c \sqrt 3c )$ = $21\times 5c (\sqrt 3c)$ = $105c \sqrt 3c$

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