How do you simplify #sqrt(135b^2c^3d) • sqrt(5b^2d)#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer Martin C. Mar 22, 2018 #15b^2cdsqrt(3c)# Explanation: #sqrt(135b^2c^3d) * sqrt(5b^2d)=sqrt(5*5c^2d^2b^4*27c)# #=5*b^2cd*sqrt(27c)=5*b^2cd*sqrt(3*3*3c)=15b^2cdsqrt(3c)# 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 1405 views around the world You can reuse this answer Creative Commons License