How do you simplify #(4+2sqrt2)(5+3sqrt2)#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer GiĆ³ Jun 4, 2015 You can multiply the two brackets as: #=(4*5)+(4*3sqrt(2))+(5*2sqrt(2))+(2*3sqrt(2)sqrt(2))=# #=20+12sqrt(2)+10sqrt(2)+12=# #=32+22sqrt(2)# #=2(16+11sqrt(2))# 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 1259 views around the world You can reuse this answer Creative Commons License