## College Algebra 7th Edition

$x=-\frac{1}{2}$
$\displaystyle \frac{1}{x^{3}}+\frac{4}{x^{2}}+\frac{4}{x}=0$ We multiply through by $x^3$: $1+4x+4x^{2}=0$ Now we factor and solve: $(1+2x)^{2}=0$ $1+2x=0$ $2x=-1$ $x=-\frac{1}{2}$