How do you simplify # 7sqrt2 · 4sqrt6#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer Tazwar Sikder Sep 18, 2016 #56 sqrt(3)# Explanation: We have: #7 sqrt(2) cdot 4 sqrt(6)# Let's express #sqrt(6)# as #sqrt(2) cdot sqrt(3)#: #= 7 cdot sqrt(2) cdot 4 cdot sqrt(2) cdot sqrt(3)# #= 7 cdot 2 cdot 4 cdot sqrt(3)# #= 56 sqrt(3)# 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 898 views around the world You can reuse this answer Creative Commons License