## College Algebra (6th Edition)

We need to factor the denominator first in order to assist in determining what the domain is: $\frac{x+5}{(x+5)(x-5)}$ Note that the $x+5$ term cancels; even so, since it is still in the denominator initially, it still remains a domain restriction. Set each term in the denominator equal to 0 in order to ascertain the domain restrictions: $x + 5 = 0$ and $x - 5 = 0$ $x = -5$ and $x = 5$ Therefore, -5 and 5 are numbers restricted from the domain, as they make the denominator zero.