How do you simplify #7sqrt108#? Algebra Radicals and Geometry Connections Simplification of Radical Expressions 1 Answer Suryin =) Oct 17, 2015 #42sqrt3# Explanation: get the #sqrt108# = #sqrt((9)(4)(3)# #9# and #4# are perfect square = #(3)(2)sqrt3# = #6sqrt3# then multiply #7# = #(7)(6)sqrt3# = #42sqrt3# Answer link Related questions How do you simplify radical expressions? How do you simplify radical expressions with fractions? How do you simplify radical expressions with variables? What are radical expressions? How do you simplify #root{3}{-125}#? How do you write # ""^4sqrt(zw)# as a rational exponent? How do you simplify # ""^5sqrt(96)# How do you write # ""^9sqrt(y^3)# as a rational exponent? How do you simplify #sqrt(75a^12b^3c^5)#? How do you simplify #sqrt(50)-sqrt(2)#? See all questions in Simplification of Radical Expressions Impact of this question 1519 views around the world You can reuse this answer Creative Commons License