How do you multiply #(3sqrtx^5)(2sqrtx^3)#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer Aviv S. Mar 3, 2018 The answer is #6x^4#. Explanation: You can rewrite #sqrtx^5# as #x^(5/2)#, and #sqrtx^3# as #x^(3/2)#: #color(white)=(3sqrtx^5)(2sqrtx^3)# #=(3x^(5/2))(2x^(3/2))# #=3*2*x^(5/2)*x^(3/2)# #=6*x^(5/2)*x^(3/2)# #=6*x^(5/2+3/2)# #=6*x^(8/2)# #=6*x^4# The answer is #6x^4#. 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 876 views around the world You can reuse this answer Creative Commons License