## Intermediate Algebra (12th Edition)

$-\dfrac{7r\sqrt{2rs}}{s^3}$
$\bf{\text{Solution Outline:}}$ To rationalize the given radical expression, $-\sqrt{\dfrac{98r^3}{s^5}} ,$ 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{98r^3}{s^5}\cdot\dfrac{s}{s}} \\\\= -\sqrt{\dfrac{98r^3s}{s^6}} .\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{49r^2\cdot2rs}{s^6}} \\\\= -\sqrt{\left(\dfrac{7r}{s^3}\right)^2\cdot2rs} .\end{array} Extracting the root of the factor that is a perfect power of the index results to \begin{array}{l}\require{cancel} -\dfrac{7r}{s^3}\sqrt{2rs} \\\\= -\dfrac{7r\sqrt{2rs}}{s^3} .\end{array}