Algebra and Trigonometry 10th Edition

Solutions: $x=0$, $x=\frac{5}{3}$ and $x=-\frac{5}{3}$
$36x^3-100x=0$ $4x(9x^2-25)=0$ $4x=0$ $x=0$ or $9x^2-25=0$ $9x^2=25$ $x^2=\frac{25}{9}$ $x=±\frac{5}{3}$ Testing: $x=0$: $36(0)^3-100(0)=0-0=0$ $x=\frac{5}{3}$: $36(\frac{5}{3})^3-100(\frac{5}{3})=36~\frac{125}{27}-\frac{500}{3}=\frac{4500}{27}-\frac{500}{3}=\frac{500}{3}-\frac{500}{3}=0$ $x=-\frac{5}{3}$: $36(-\frac{5}{3})^3-100(-\frac{5}{3})=36~(-\frac{125}{27})+\frac{500}{3}=-\frac{4500}{27}+\frac{500}{3}=-\frac{500}{3}+\frac{500}{3}=0$