How do you simplify #(sqrt5-2)(sqrt5-1)#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer Barney V. Apr 23, 2017 #color(blue)(=7-3sqrt5# Explanation: #(sqrt5-2)(sqrt5-1)# #color(white)(aaaaaaaaaaaaa)##sqrt5-2# #color(white)(aaaaaaaaaaaaa)## sqrt5-1# #color(white)(aaaaaaaaaaaaa)##------# #color(white)(aaaaaaaaaaaaaaa)##5-2sqrt5# #color(white)(aaaaaaaaaaaaaaaaaa)##-sqrt5+2# #color(white)(aaaaaaaaaaaaa)##------# #color(white)(aaaaaaaaaaaaaaa)##color(blue)(5-3sqrt5+2# #color(blue)(=7-3sqrt5# 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 364 views around the world You can reuse this answer Creative Commons License