How do you simplify #sqrt14sqrt35#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer elaf · Evan May 26, 2018 7#sqrt10# Explanation: #sqrt14 = sqrt(7*2)# #sqrt35 =sqrt(5*7)# Hence, #sqrt14*sqrt35=sqrt7*sqrt2*sqrt5*sqrt7# Simplify, #sqrt14*sqrt35=7sqrt10# 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 2061 views around the world You can reuse this answer Creative Commons License