How do you multiply #(3 + sqrt3) (4 + 3sqrt3)#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer Meave60 Apr 30, 2015 The answer is #21+13sqrt3# Use the FOIL method. #(3+sqrt3)(4+3sqrt3)# = #3*4+3*3sqrt3+4sqrt3+3sqrt3sqrt3# = #12+9sqrt3+4sqrt3+3*3# = #12+13sqrt3+9# = #21+13sqrt3# 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 1222 views around the world You can reuse this answer Creative Commons License