## College Algebra (6th Edition)

$\frac{4x+16}{(x+3)^{2}}; x \ne -3;$ or $\frac{4(x+4)}{(x+3)^{2}}; x \ne -3;$
$\frac{4}{x^{2}+6x+9} + \frac{4}{x+3}$ $x^{2}+6x+9$ is in the form of $A^{2}+2AB+B^{2} = (A+B)^{2}$ $x^{2}+6x+9 = (x+3)^{2}$ So, $=\frac{4}{ (x+3)^{2}} + \frac{4}{x+3}$ Taking Least Common Denominator, $=\frac{4+4(x+3)}{ (x+3)^{2}} ; x \ne -3$ $=\frac{4+4x+12}{ (x+3)^{2}} ; x \ne -3$ $= \frac{4x+16}{(x+3)^{2}}; x \ne -3;$ $=\frac{4(x+4)}{(x+3)^{2}}; x \ne -3;$