## Intermediate Algebra (6th Edition)

$b. (x+1)^{2}$
The lowest common denominator for this problem can be found by comparing the three denominators. 1. $(x+1)$ 2. $(x+1)(x+1)$ 3. $(x+1)$ If we multiply the first and third denominators by $(x+1)$ we will arrive at a LCD of $(x+1)^{2}$.