How do you simplify #10sqrt6 *sqrt2#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer Don't Memorise May 15, 2015 #color(blue)(sqrta * sqrt b = sqrt (ab)# Hence #sqrt6*sqrt2 = sqrt(6xx2) = color(blue)(sqrt(12)# # 10 sqrt12 = 10sqrt(3xx4)# (# sqrt4 =2#) #10sqrt12= 10xx2sqrt3# #= 20sqrt3# 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 1235 views around the world You can reuse this answer Creative Commons License