How do you simplify #sqrt2 (sqrt8 + sqrt4)#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer Taner Müftüoğlu Sep 25, 2016 multiply and try to get rid of #sqrt()#'s if possible Explanation: #sqrt(2)(sqrt(8) + sqrt(4)) = sqrt(2)sqrt(8) + sqrt(2)sqrt(4)# #=sqrt(16) + sqrt(2)*2 # #= 4+2sqrt(2)# or #= 4+sqrt(8) # 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 1065 views around the world You can reuse this answer Creative Commons License