1. **State the problem:** Solve the inequality $$\frac{(x - 1)(x + 6)}{x + 2} > 0$$ and express the solution in interval notation. 2. **Identify critical points:** The numerator is zero when $x - 1 = 0$ or $x + 6 = 0$, so $x = 1$ or $x = -6$. The denominator is zero when $x + 2 = 0$, so $x = -2$. These points divide the number line into intervals. 3. **Intervals to test:** $(-\infty, -6)$, $(-6, -2)$, $(-2, 1)$, and $(1, \infty)$. 4. **Test each interval:** - For $x < -6$, pick $x = -7$: numerator $(-7-1)(-7+6) = (-8)(-1) = 8 > 0$, denominator $-7+2 = -5 < 0$, so fraction is $\frac{+}{-} = -$ (negative). - For $-6 < x < -2$, pick $x = -4$: numerator $(-4-1)(-4+6) = (-5)(2) = -10 < 0$, denominator $-4+2 = -2 < 0$, fraction is $\frac{-}{-} = +$ (positive). - For $-2 < x < 1$, pick $x = 0$: numerator $(0-1)(0+6) = (-1)(6) = -6 < 0$, denominator $0+2 = 2 > 0$, fraction is $\frac{-}{+} = -$ (negative). - For $x > 1$, pick $x = 2$: numerator $(2-1)(2+6) = (1)(8) = 8 > 0$, denominator $2+2 = 4 > 0$, fraction is $\frac{+}{+} = +$ (positive). 5. **Determine where fraction is positive:** On intervals $(-6, -2)$ and $(1, \infty)$. 6. **Check critical points:** - At $x = -6$ numerator zero, fraction zero, inequality is strict $>$, so exclude $-6$. - At $x = -2$ denominator zero, undefined, exclude $-2$. - At $x = 1$ numerator zero, fraction zero, exclude $1$. 7. **Final solution in interval notation:** $$(-6, -2) \cup (1, \infty)$$