How do you simplify #(sqrt2 *sqrt8)/2#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer Elvan Burcu K. Apr 9, 2015 You can multiply #sqrt2# and #sqrt8# into a one squareroot. So the solution will be; #(sqrt2 . sqrt8) / 2 # = #sqrt16 /2 # = #4/2# = 2 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 1455 views around the world You can reuse this answer Creative Commons License