How do you simplify #sqrt(2c^2)*sqrt(8c)#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer Hriman Apr 16, 2018 #4|c|sqrtc# Explanation: Note that: #sqrta*sqrtb= sqrt(ab)# So: #sqrt(2c^2)(8c)= sqrt(2c^2*8c)# Therefore: #sqrt(16c^3)# #sqrt(16c^2*c)# #4|c|sqrtc# Answer link Related questions How do you simplify #\frac{2}{\sqrt{3}}#? How do you multiply and divide radicals? How do you rationalize the denominator? What is Multiplication and Division of Radicals? How do you simplify #7/(""^3sqrt(5)#? How do you multiply #(sqrt(a) +sqrt(b))(sqrt(a)-sqrt(b))#? How do you rationalize the denominator for #\frac{2x}{\sqrt{5}x}#? Do you always have to rationalize the denominator? How do you simplify #sqrt(5)sqrt(15)#? How do you simplify #(7sqrt(13) + 2sqrt(6))(2sqrt(3)+3sqrt(6))#? See all questions in Multiplication and Division of Radicals Impact of this question 1359 views around the world You can reuse this answer Creative Commons License