How do you simplify #(2sqrt7+4)^2#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer bp · Ahmet C Apr 13, 2015 #44+ 16sqrt7# To simplify use the algebraic identity #(a+b)^2#= #a^2+2ab +b^2#. It would give #(2sqrt7)^2 +2 (2sqrt7)(4) + (4)^2 = 44+16sqrt7# 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 1396 views around the world You can reuse this answer Creative Commons License