Elementary Algebra

$-\frac{8}{x^{2}}$
Since all terms are multiplied by each other, we can collect constants and variables separately in their own parentheses. We then use the rule $a^{m}\times a^{n}=a^{m+n}$ and $\frac{1}{x^{2}}=x^{-2}$ to simplify the expression: $(-4x^{-5})(2x^{3})$ =$(\frac{-4}{x^{5}})(2x^{3})$ =$(-4\times2)\times(\frac{x^{3}}{x^{5}})$ =$(-8)\times x^{3-5}$ =$(-8)\times x^{-2}$ =$\frac{-8}{x^{2}}$ =$-\frac{8}{x^{2}}$