## Algebra 1: Common Core (15th Edition)

$\frac{1}{7x}$
The excluded values are x=0 because it makes the denominator zero. Reduce $\frac{4x^{3}}{28x^{4}}$ by dividing by $\frac{4}{4}$. $\frac{4x^{3}}{28x^{4}}$ $\div$ $\frac{4}{4}$ = $\frac{x^{3}}{7x^{4}}$. Subtract the powers of x. $\frac{x^{3}}{7x^{4}}$ = $\frac{1}{7x^{4-3}}$ = $\frac{1}{7x}$