How do you simplify #sqrt2(sqrt3-sqrt5)#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer David L. Apr 5, 2017 #sqrt6 - sqrt10# Explanation: Multiply #sqrt2# through the brackets #sqrt2*sqrt3 = sqrt6# and #sqrt2*sqrt5 = sqrt10#** #(sqrt2*sqrt3)-(sqrt2*sqrt5)# #sqrt6 - sqrt10# 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 1656 views around the world You can reuse this answer Creative Commons License