How do you multiply # (x^(3/2) + 2/sqrt3)^2#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer smendyka · JS Jun 9, 2017 See a solution process below: Explanation: This is a special form of quadratic: #(a + b)^2 = a^2 + 2ab + b^2# Substituting #x^(3/2)# for #a# and #2/sqrt(3)# for #b# gives: #(x^(3/2) + b)^2 = (x^(3/2))^2 + (2 * x^(3/2) * 2/sqrt(3)) + (2/sqrt(3))^2 =# #x^3 + (4x^(3/2))/sqrt(3) + 4/3# 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 384 views around the world You can reuse this answer Creative Commons License