How do you multiply #sqrt2(3sqrt14 - sqrt7)#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer Serena D. Mar 27, 2018 #6sqrt7-sqrt14# Explanation: Multiply #sqrt2# by each term in the parentheses #sqrt2(3sqrt14-sqrt7)# #sqrt2*3sqrt14-sqrt2*sqrt7# #3sqrt28-sqrt14 rarr 3sqrt28# can be simplified #3sqrt28=3sqrt4*sqrt7 rarr# #4# is a perfect square #6*sqrt7=6sqrt7=3sqrt28# #6sqrt7-sqrt14# 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 1415 views around the world You can reuse this answer Creative Commons License