How do you simplify #sqrt(32xy^2) div sqrt4#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer Evan Mar 11, 2018 #ysqrt(8x)# Explanation: #sqrt(32xy^2)-:sqrt4# #=(sqrt32xxsqrtx xxsqrt(y^2))-:sqrt4# #=(cancel(sqrt4)xxsqrt8xxsqrtx xxy)-:cancel(sqrt4# #=yxxsqrt8xxsqrtx# #=ysqrt(8x)# 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 1350 views around the world You can reuse this answer Creative Commons License