How do you simplify #(2sqrt5+3sqrt2)^2#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals ↳Redirected from "How can molarity be used as a conversion factor?" 1 Answer Tony B Jul 5, 2016 #38+12sqrt(10)# Explanation: #color(blue)((2sqrt(5)+3sqrt(2))color(brown)((2sqrt(5)+3sqrt(2))# #color(brown)(color(blue)(2sqrt(5)color(brown)((2sqrt(5)+3sqrt(2))color(blue)(" "+" "3sqrt(2)color(brown)((2sqrt(5)+3sqrt(2))# #(4xx5)+6sqrt(5)sqrt(2)" "+" "6sqrt(5)sqrt(2)+(9xx2)# #20+12sqrt(5)sqrt(2)+18# But #sqrt(5)sqrt(2)" is the same as "sqrt(5xx2)= sqrt(10)# #20+18+12sqrt(10)# #38+12sqrt(10)# 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 1296 views around the world You can reuse this answer Creative Commons License