How do you multiply #sqrt2 *sqrt8#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer George C. May 10, 2015 #sqrt(2)sqrt(8) = sqrt(2*8) = sqrt(16) = 4# Based on #sqrt(a)sqrt(b) = sqrt(ab)#. 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 1075 views around the world You can reuse this answer Creative Commons License