How do you divide #2sqrt6# by #sqrt2# and leave your answer in radical form? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer Alan P. Mar 22, 2015 Note that #sqrt(6)/sqrt(2) = sqrt(6/2)# So #(2sqrt(6))/sqrt(2) = 2 sqrt(6/2) = 2 sqrt(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 #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 1416 views around the world You can reuse this answer Creative Commons License