The critical points are x = 3 (from the numerator) and x = -5 (from the denominator). Note that x cannot be -5 because it would make the denominator zero. Testing intervals: for x 0; for -5 < x < 3 (e.g., 0), (-3)/(5) 3 (e.g., 4), (1)/(9) > 0. The inequality ≤ 0 is satisfied in the interval (-5, 3]. We include 3 because of the 'equal to' part of the inequality, but exclude -5 because the expression is undefined there.