How do you simplify #sqrt(3x)((sqrt6x)-sqrt12)#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer Surya K. Nov 14, 2017 #3xsqrt(2)-6# Explanation: Multiply #sqrt(3x)# with the rest of the terms: #sqrt(3x)((sqrt(6x))-sqrt(12))# #sqrt(3x)*sqrt(6x))-sqrt(3x)*sqrt(12))# #sqrt(3*6)*sqrt(x*x)-sqrt(3*12)# #sqrt(18)*sqrt(x^2)-sqrt(36)# #3sqrt(2)*x-6# #3xsqrt(2)-6# 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 1092 views around the world You can reuse this answer Creative Commons License