## Introductory Algebra for College Students (7th Edition)

$4y^{100}\sqrt{2}$
Getting the perfect factors and then extracting its root, the simplified form of the given expression, $\sqrt{32y^{200}} ,$ is \begin{array}{l}\require{cancel} \sqrt{16y^{200}\cdot2} \\\\= \sqrt{(4y^{100})^2\cdot2} \\\\= 4y^{100}\sqrt{2} .\end{array}