How do you simplify #4 / sqrt 5#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer Alan P. Sep 29, 2015 #4/sqrt(5) = 4/5sqrt(5)# #color(white)("XXXXXXXX")# which some would consider a "simplification". Explanation: #4/sqrt(5) = 4/sqrt(5) * sqrt(5)/sqrt(5) = (4sqrt(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 1163 views around the world You can reuse this answer Creative Commons License