Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$\sqrt{15}+4\sqrt{6}-6\sqrt{10}-48$
Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the given expression is equivalent to \begin{array}{l}\require{cancel} (\sqrt{3}-3\sqrt{8})(\sqrt{5}+2\sqrt{8}) \\\\= \sqrt{3}(\sqrt{5})+\sqrt{3}(2\sqrt{8})-3\sqrt{8}(\sqrt{5})-3\sqrt{8}(2\sqrt{8}) .\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} \sqrt{3}(\sqrt{5})+\sqrt{3}(2\sqrt{8})-3\sqrt{8}(\sqrt{5})-3\sqrt{8}(2\sqrt{8}) \\\\= \sqrt{3(5)}+2\sqrt{3(8)}-3\sqrt{8(5)}-6(\sqrt{8})^2 \\\\= \sqrt{15}+2\sqrt{24}-3\sqrt{40}-6(8) \\\\= \sqrt{15}+2\sqrt{24}-3\sqrt{40}-48 .\end{array} Extracting the factors that are perfect powers of the index, the expression above simplifies to \begin{array}{l}\require{cancel} \sqrt{15}+2\sqrt{24}-3\sqrt{40}-48 \\\\= \sqrt{15}+2\sqrt{4\cdot6}-3\sqrt{4\cdot10}-48 \\\\= \sqrt{15}+2\sqrt{(2)^2\cdot6}-3\sqrt{(2)^2\cdot10}-48 \\\\= \sqrt{15}+2(2)\sqrt{6}-3(2)\sqrt{10}-48 \\\\= \sqrt{15}+4\sqrt{6}-6\sqrt{10}-48 .\end{array}