How do you simplify #(sqrtm-sqrt5)^2#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer Isini M. Jun 17, 2017 #=m-2sqrt(5m)+5# Explanation: Recall that #(a-b)^2=a^2-2ab+b^2# #(sqrt(m)-sqrt(5))^2# #=(sqrt(m))^2-2(sqrt(m))(sqrt(5))+(sqrt(5))^2# #=m-2sqrt(m×5)+5# #=m-2sqrt(5m)+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 1163 views around the world You can reuse this answer Creative Commons License