How do you divide #(16sqrt40)/(4sqrt5)#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer Joe D. Apr 2, 2015 #(16/4)*(sqrt(40)/sqrt(5)) = # #4 * sqrt(8*5)/sqrt(5) = # #4 * (sqrt(8) * sqrt(5))/sqrt(5) = # #4 * sqrt(2^2 * 2) = # #4 * 2 * sqrt(2) = 8sqrt(2)# 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 1568 views around the world You can reuse this answer Creative Commons License