How do you multiply #6(sqrt3+5sqrt2)#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer Alan P. Mar 24, 2015 This is a straight application of the distributive property: #6(sqrt(3) + 5sqrt(2))# #= (6)(sqrt(3)) + (6)(5sqrt(2))# #= 6sqrt(3) + 30sqrt(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 1283 views around the world You can reuse this answer Creative Commons License