## Intermediate Algebra (12th Edition)

$52+6\sqrt{35}$
$\bf{\text{Solution Outline:}}$ To simplify the given radical expression, $(3\sqrt{5}+2\sqrt{7})^2 ,$ use the special product on squaring binomials and the laws of radicals. $\bf{\text{Solution Details:}}$ Using the square of a binomial which is given by $(a+b)^2=a^2+2ab+b^2$ or by $(a-b)^2=a^2-2ab+b^2,$ the expression above is equivalent to \begin{array}{l}\require{cancel} (3\sqrt{5})^2+2(3\sqrt{5})(\sqrt{7})+(\sqrt{7})^2 \\\\= 9(5)+6(\sqrt{5})(\sqrt{7})+7 \\\\= 45+6(\sqrt{5})(\sqrt{7})+7 \\\\= 52+6(\sqrt{5})(\sqrt{7}) .\end{array} Using the Product Rule of radicals which is given by $\sqrt[m]{x}\cdot\sqrt[m]{y}=\sqrt[m]{xy},$ the expression above is equivalent to\begin{array}{l}\require{cancel} 52+6\sqrt{5(7)} \\\\= 52+6\sqrt{35} .\end{array}