How do you simplify #(5x + sqrt7)^2#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer LM Apr 1, 2017 #25x^2+10xsqrt7+7# Explanation: #(5x+sqrt7)^2 = (5x+sqrt7)(5x+sqrt7)# expand brackets using FOIL: #(5x+sqrt7)^2 =(5x+sqrt7)(5x+sqrt7)# #= 25x^2+5xsqrt7+5xsqrt7+sqrt7^2# #=25x^2+10xsqrt7+7# 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 830 views around the world You can reuse this answer Creative Commons License