How do you simplify #(3n^2+sqrt(2n^2))/sqrt(10n)#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer Mark D. Jun 4, 2018 #[3sqrt(10)n^2+2sqrt5n]/10# Explanation: Mutiply by #[sqrt10]/[sqrt10]# #[3n^2+sqrt(2n^2)]/[sqrt(10n)]xx[sqrt10]/[sqrt10]# #[3sqrt(10)n^2+sqrt(20n^2)]/sqrt(100)# #[3sqrt(10)n^2+2sqrt5n]/10# 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 2019 views around the world You can reuse this answer Creative Commons License