How do you simplify #sqrt(7)*(2 sqrt(3) + 3 sqrt(7))#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer KillerBunny Oct 16, 2015 #2sqrt(21)+21#, or if you prefer #sqrt(21)*(2+sqrt(21))#. Explanation: Expand the multiplications: #sqrt(7)*(2sqrt(3)+3*sqrt(7)) = 2sqrt(3*7) + 3sqrt(7*7)# Of course, #sqrt(7*7)=sqrt(49)=7#, so the expression begins #2sqrt(21)+3*7=2sqrt(21)+21#. Since #21=sqrt(21)*sqrt(21)#, you can factor #sqrt(21)# and obtain #sqrt(21)*(2+sqrt(21))# 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 1548 views around the world You can reuse this answer Creative Commons License