How do you multiply #(2sqrt7+sqrt5)(sqrt3+sqrt2)(2sqrt7-sqrt5)#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer George C. Jun 5, 2015 It's easiest to multiply #(2sqrt(7)+sqrt(5))(2sqrt(7)-sqrt(5))# first... #(2sqrt(7)+sqrt(5))(2sqrt(7)-sqrt(5))# is of the form #(a+b)(a-b) = a^2 - b^2# with #a=2sqrt(7)# and #b=sqrt(5)# So: #(2sqrt(7)+sqrt(5))(2sqrt(7)-sqrt(5))# #=(2sqrt(7))^2-sqrt(5)^2 = (4*7)-5 = 28 - 5 = 23# Then #(2sqrt(7)+sqrt(5))(sqrt(3)+sqrt(2))(2sqrt(7)-sqrt(5))# #= (2sqrt(7)+sqrt(5))(2sqrt(7)-sqrt(5))(sqrt(3)+sqrt(2))# #=23(sqrt(3)+sqrt(2)) = 23sqrt(3)+23sqrt(2)# 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 1316 views around the world You can reuse this answer Creative Commons License