How do you simplify #sqrt5(6-sqrt5)#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer Nam D. Apr 24, 2018 #6sqrt5-5# Explanation: Given: #sqrt5(6-sqrt5)# Using the distributive property, #a(b-c)=ab-ac#. #:.=6sqrt(5)-sqrt(5)*sqrt(5)# #=6sqrt5-(sqrt(5))^2# #=6sqrt(5)-5# 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 918 views around the world You can reuse this answer Creative Commons License