How do you simplify #sqrt2(sqrt6+sqrt8)+sqrt3#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer Tessalsifi Jun 7, 2015 We are going to expand : #=sqrt2*sqrt6+sqrt2*sqrt8+sqrt3# #=sqrt12+sqrt16+sqrt3# #=sqrt(4*3)+4+sqrt3# ( since #sqrt16=4# ) #=sqrt4*sqrt3+4+sqrt3# #=2sqrt3+4+sqrt3# ( since #sqrt4=2# ) #=3sqrt3+4# 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 1641 views around the world You can reuse this answer Creative Commons License