How do you simplify #sqrt21*sqrt14##? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer Bill Jorgensen Jun 9, 2018 #7sqrt6# Explanation: #sqrt21*sqrt14# Use the rule: #sqrta * sqrtb = sqrt(a*b)# #sqrt21*sqrt14 = sqrt(21*14)=sqrt294=sqrt(7*7*6)# #=sqrt(7^2*6)=7sqrt6# 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 1342 views around the world You can reuse this answer Creative Commons License