How do you multiply # (3sqrt6)(2sqrt50)#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer Don't Memorise Dec 21, 2015 #=color(blue)(60sqrt3# Explanation: #color(blue)(3sqrt6) xx 2sqrt50# #= color(blue)(3 xx2) xx sqrt6 xx sqrt50# #= color(blue)(6) xx sqrt300# Prime factorising #300# #300=2*2*5*5*3# So, #sqrt300=sqrt(2*2*5*5*3)# #=sqrt(2^2xx5^2xx3)# #=2 xx 5sqrt( 3)# #=10sqrt( 3)# The expression becomes: #= color(blue)(6) xx sqrt300 = color(blue)(6 xx 10sqrt3# #=color(blue)(60sqrt3# 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 904 views around the world You can reuse this answer Creative Commons License