How do you simplify #-4sqrt15*-sqrt3#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer Nityananda Jun 3, 2017 #12sqrt5# Explanation: Given, #-4sqrt15* -sqrt3 = (-xx-)4sqrt(3*5)*sqrt3# #rArr +4sqrt3*4sqrt5*sqrt3# #rArr 4sqrt3sqrt3*sqrt5# #rArr 4*3^(1/2)3^(1/2)*sqrt5# #rArr 4*3^(1/2+1/2)*sqrt5# #rArr 4*3^(2/2)*sqrt5# #rArr 4*3*sqrt5 = 12sqrt5# 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 921 views around the world You can reuse this answer Creative Commons License