## Intermediate Algebra (12th Edition)

$\dfrac{2\sqrt{5rm}}{m^2}$
$\bf{\text{Solution Outline:}}$ To rationalize the given radical expression, $\dfrac{2\sqrt{5r}}{\sqrt{m^3}} ,$ 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} \dfrac{2\sqrt{5r}}{\sqrt{m^3}}\cdot\dfrac{\sqrt{m}}{\sqrt{m}} .\end{array} Using the Product Rule of radicals which is given by $\sqrt[m]{x}\cdot\sqrt[m]{y}=\sqrt[m]{xy},$ the expression above is equivalent to\begin{array}{l}\require{cancel} \dfrac{2\sqrt{5rm}}{\sqrt{m^3(m)}} \\\\= \dfrac{2\sqrt{5rm}}{\sqrt{m^4}} \\\\= \dfrac{2\sqrt{5rm}}{\sqrt{(m^2)^2}} .\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\sqrt{5rm}}{m^2} .\end{array}