## Intermediate Algebra (12th Edition)

$-\dfrac{5m^2\sqrt{6mn}}{n^2}$
$\bf{\text{Solution Outline:}}$ To rationalize the given radical expression, $-\sqrt{\dfrac{150m^5}{n^3}} ,$ multiply the radicand by an expression equal to $1$ which will make the denominator a perfect power of the index. $\bf{\text{Solution Details:}}$ Multiplying the radicand 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{\dfrac{150m^5}{n^3}\cdot\dfrac{n}{n}} \\\\= -\sqrt{\dfrac{150m^5n}{n^4}} .\end{array} Writing the radicand as an expression that contains a factor that is a perfect power of the index results to \begin{array}{l}\require{cancel} -\sqrt{\dfrac{25m^4}{n^4}\cdot6mn} \\\\= -\sqrt{\left( \dfrac{5m^2}{n^2} \right)^2\cdot6mn} .\end{array} Extracting the root of the factor that is a perfect power of the index results to \begin{array}{l}\require{cancel} -\dfrac{5m^2}{n^2}\sqrt{6mn} \\\\= -\dfrac{5m^2\sqrt{6mn}}{n^2} .\end{array}