## Precalculus (6th Edition)

$\color{blue}{\dfrac{25p^2}{9}}$
Use the rule $\dfrac{a}{b} \div \dfrac{c}{d} = \dfrac{a}{b} \times \dfrac{d}{c}$ to obtain: $\dfrac{15p^3}{9p^2} \times \dfrac{10p^2}{6p}$ Cancel common factors, then multiply to obtain: $\require{cancel} =\dfrac{\cancel{15}^5p^3}{\cancel{9}^3\cancel{p^2}} \times \dfrac{10\cancel{p^2}}{6p} \\=\dfrac{5p^3}{3} \times \dfrac{10}{6p} \\=\dfrac{5\cancel{p^3}p^2}{3} \times \dfrac{\cancel{10}5}{\cancel{6}3\cancel{p}} \\=\color{blue}{\dfrac{25p^2}{9}}$