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

$2xy^{2}\sqrt[4]{xy^3}$
Extracting the factors that are perfect roots of the given index, the given expression, $\sqrt[4]{16x^5y^{11}} ,$ simplifies to \begin{array}{l}\require{cancel} \sqrt[4]{16x^4y^{8}\cdot xy^3} \\\\= \sqrt[4]{(2xy^{2})^4\cdot xy^3} \\\\= 2xy^{2}\sqrt[4]{xy^3} \end{array} * Note that it is assumed that no radicands were formed by raising negative numbers to even powers.