How do you multiply #(sqrt10 - 9)^2# and write the product in simplest form? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer GiĆ³ Mar 24, 2015 You can write (considering that #(a-b)^2=a^2-2ab+b^2#): #(sqrt(10)-9)*(sqrt(10)-9)=(sqrt(10))^2-2*9*sqrt(10)+9^2=# #=10-18sqrt(10)+81=# #=91-18sqrt(10)=# 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 1538 views around the world You can reuse this answer Creative Commons License