## Precalculus (6th Edition) Blitzer

$\displaystyle \frac{4(x+4)}{(x+3)^{2}},\quad x\neq-3$
Factor each denominator. Recognize a perfect square in the first. $\displaystyle \frac{4}{x^{2}+6x+9}+\frac{4}{x+3}=\frac{4}{(x+3)^{2}}+\frac{4}{x+3}=...$ The denominators are different. Find a common denominator. LCD=$(x+3)^{2}$ $=\displaystyle \frac{4}{(x+3)^{2}}+\frac{4}{x+3}\times\frac{x+3}{x+3}$ $=\displaystyle \frac{4}{(x+3)^{2}}+\frac{4x+12}{(x+3)^{2}}$ $=\displaystyle \frac{4+4x+12}{(x+3)^{2}}$ $=\displaystyle \frac{4x+16}{(x+3)^{2}}$ = $\displaystyle \frac{4(x+4)}{(x+3)^{2}},\quad x\neq-3$