How do you simplify #sqrt12 times sqrt27#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer Gerardina C. Jun 23, 2016 18 Explanation: You know that #sqrt(12)=sqrt(2^2*3)=2sqrt(3)# and #sqrt(27)=sqrt(3^3)=3sqrt(3)# so you can sum them: #sqrt(12)*sqrt(27)=2sqrt(3)*3sqrt(3)=6sqrt(3)^2=6*3=18# 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 815 views around the world You can reuse this answer Creative Commons License