How do you simplify #(sqrt2 * sqrt8)/2#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 2 Answers Deevona Apr 9, 2015 #sqrt(2)# #*# 2 #sqrt(2)# #-:# 2 = = 2 Answer link Don't Memorise Apr 9, 2015 #(sqrt2 * sqrtcolor(blue)(8))/2# #= (sqrt2 * sqrtcolor(blue)(2^3))/2 # (We write 8 as #2^3#) #= (sqrt2 * sqrtcolor(blue)(2^2*2))/2 # (We write #2^3# as #2^2*2#) #= (sqrt2 * (cancel2*sqrt2))/cancel2# #(sqrt(2^2*2) =sqrt(2^2)*sqrt2)# #= sqrt2 * sqrt2 # #color(green)(= 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 1648 views around the world You can reuse this answer Creative Commons License