How do you simplify #sqrt(3x^3) * sqrt(6x^2)#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer George C. Sep 3, 2015 #sqrt(3x^3)*sqrt(6x^2) = sqrt(18x^5)=3xsqrt(2x)# (assuming #x >= 0#) Explanation: If #sqrt(a)# and #sqrt(b)# are Real, then #a, b >= 0# and #sqrt(a)sqrt(b) = sqrt(ab)# In our case, if both #sqrt#'s are Real, then: #sqrt(3x^3)*sqrt(6x^2) = sqrt(3x^3*6x^2) = sqrt(18x^5) = sqrt((3x^2)^2*2x)# #= sqrt((3x^2)^2)sqrt(2x) = 3x^2sqrt(2x)# 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 1206 views around the world You can reuse this answer Creative Commons License