How do you simplify #(3*sqrt12)(sqrt6)#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer GiĆ³ Jun 23, 2015 I found; #18*sqrt(2)# Explanation: I would write it as: #(3*sqrt(4*3))(sqrt(6))=(3*2*sqrt(3))(sqrt(6))=# #=6*sqrt(3)*sqrt(6)=6*sqrt(18)=6*sqrt(9*2)=# #=6*3*sqrt(2)=18*sqrt(2)# 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 1201 views around the world You can reuse this answer Creative Commons License