How do you simplify #(4sqrt6)/ sqrt30#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer GiĆ³ Mar 31, 2015 You can write: #4sqrt(6)/(sqrt(30))=4sqrt(6)/(sqrt(6*5))=4sqrt(6)/(sqrt(6)sqrt(5))=# #=4/sqrt(5)=# rationalizing you get: #=4/sqrt(5)*sqrt(5)/sqrt(5)=4sqrt(5)/5# 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 1499 views around the world You can reuse this answer Creative Commons License