## College Algebra (6th Edition)

$\frac{7(x+1)^{2}}{2x^{2}}; x \ne -5,-1,0,1,5$
The given expression can be factored as $=\frac{x(x^{2}-25)}{4x^{2}} \times \frac{2(x^{2}-1)}{(x-5)(x-1)} \div \frac{x(x+5)}{7(x+1)} ; x \ne -5,-1,0,1,5$ $(x^{2}-25)$ and $(x^{2}-1)$ are difference of squares, so we can further factor as $= \frac{x(x+5)(x-5)}{4x^{2}} \times \frac{2(x-1)(x+1)}{(x-5)(x-1)} \div \frac{x(x+5)}{7(x+1)} ; x \ne -5,-1,0,1,5$ $x \ne -5,-1,0,1,5$, otherwise they make the denominators zero. $= \frac{x(x+5)(x-5)}{4x^{2}} \times \frac{2(x-1)(x+1)}{(x-5)(x-1)} \times\frac{7(x+1)}{x(x+5)} ; x \ne -5,-1,0,1,5$ Divide out common factors, the result is $=\frac{7(x+1)^{2}}{2x^{2}}; x \ne -5,-1,0,1,5$