## Precalculus (6th Edition)

$\color{blue}{\dfrac{4}{x}}$
Cancel the common factors first before performing the actual multiplication to obtain: $\require{cancel} =\dfrac{2x}{5} \cdot \dfrac{2(5)}{x(x)} \\=\dfrac{2\cancel{x}}{\cancel{5}} \cdot \dfrac{2(\cancel{5})}{x\cancel{(x)}} \\=\dfrac{2}{1} \cdot \dfrac{2}{x} \\=\color{blue}{\dfrac{4}{x}}$