How do you simplify #(-4 sqrt(6)) / sqrt(27)#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer anor277 Sep 4, 2016 #(4sqrt(6))/sqrt27# #=# #(-4sqrt2)/3# Explanation: #(4sqrt(6))/sqrt27# #=# #(-4sqrt(2)xxcancelsqrt(3))/(sqrt(9)xxcancelsqrt3)# #=# #(-4sqrt(2))/sqrt(9)# #=# #(-4sqrt2)/3# So in effect, you factorize the square roots. 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 1001 views around the world You can reuse this answer Creative Commons License