How do you simplify #(sqrt 5 - 2)(sqrt 5 - 4)#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer GiĆ³ Sep 24, 2015 I found: #13-6sqrt(5)# Explanation: You can multiply to get: #(sqrt(5)-2)(sqrt(5)-4)=sqrt(5)sqrt(5)-4sqrt(5)-2sqrt(5)+8=# #=5-6sqrt(5)+8=13-6sqrt(5)# 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 2205 views around the world You can reuse this answer Creative Commons License