How do you simplify #4 sqrt5 times 4 sqrt6#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer Peter May 12, 2016 Given #4sqrt5*4sqrt6#, We multiply 4 with 4 and #sqrt5# with #sqrt6# to get #4*4=16# #sqrt5*sqrt6=sqrt30# Therefore the equation #4sqrt5*4sqrt6# =16#sqrt30# 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 1650 views around the world You can reuse this answer Creative Commons License