How do you simplify #(6 + sqrt 5) (6 - sqrt5)#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer Alan P. Oct 15, 2015 #(6+sqrt(5))(6-sqrt(5)) = 31# Explanation: #(a+b)(a-b) = (a^2-b^2)# Replacing #a# with #6# and #b# with #sqrt(5)# #(6+sqrt(5))(6-sqrt(5))# #color(white)("XXX")=(6^2-(sqrt(5))^2)# #color(white)("XXX")=(36-5)# #color(white)("XXX")=31# 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 7773 views around the world You can reuse this answer Creative Commons License