## Intermediate Algebra (12th Edition)

$\dfrac{2\sqrt[4]{ x^3}}{x}$
$\bf{\text{Solution Outline:}}$ To rationalize the given radical expression, $\sqrt[4]{\dfrac{16}{x}} ,$ multiply the numerator and the denominator by an expression equal to $1$ which will make the denominator a perfect power of the index. $\bf{\text{Solution Details:}}$ Multiplying the numerator and the denominator by an expression equal to $1$ which will make the denominator a perfect power of the index results to \begin{array}{l}\require{cancel} \sqrt[4]{\dfrac{16}{x}\cdot\dfrac{x^3}{x^3}} \\\\= \sqrt[4]{\dfrac{16x^3}{x^4}} .\end{array} Rewriting the radicand using an expression that contains a factor that is a perfect power of the index results to \begin{array}{l}\require{cancel} \sqrt[4]{\dfrac{16}{x^4}\cdot x^3} \\\\= \sqrt[4]{\left( \dfrac{2}{x} \right)^4\cdot x^3} .\end{array} Extracting the root of the radicand that is a perfect power of the index results to \begin{array}{l}\require{cancel} \dfrac{2}{x}\sqrt[4]{ x^3} \\\\= \dfrac{2\sqrt[4]{ x^3}}{x} .\end{array}