How do you simplify #sqrt(5)sqrt(15)#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer Meave60 Feb 13, 2015 You could also do either of the following: #sqrt(5)sqrt(15) = sqrt(5)*sqrt(5*3)=sqrt(5)*sqrt(5)*sqrt(3)=sqrt(25)*sqrt(3)=5sqrt(3)# #sqrt(5)sqrt(15) = sqrt(5)*sqrt(5)*sqrt(3)=sqrt(25)*sqrt(3)=5sqrt(3)# 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 #(7sqrt(13) + 2sqrt(6))(2sqrt(3)+3sqrt(6))#? How do you simplify #(sqrt(7) + sqrt(5))^4 + (sqrt(7) - sqrt(5))^4#? See all questions in Multiplication and Division of Radicals Impact of this question 20864 views around the world You can reuse this answer Creative Commons License