1. **State the problem:** Solve the inequality $$\frac{x^2 - 6x + 8}{x + 3} < 0$$ and represent the solution on a number line. 2. **Factor the numerator:** The quadratic expression in the numerator can be factored as: $$x^2 - 6x + 8 = (x - 2)(x - 4)$$ 3. **Rewrite the inequality:** $$\frac{(x - 2)(x - 4)}{x + 3} < 0$$ 4. **Identify critical points:** These are values where the numerator or denominator is zero: - Numerator zeros: $x = 2$, $x = 4$ - Denominator zero: $x = -3$ (excluded from domain) 5. **Determine intervals:** The critical points divide the number line into four intervals: $$(-\infty, -3), (-3, 2), (2, 4), (4, \infty)$$ 6. **Test each interval:** Choose a test point in each interval and evaluate the sign of the expression. - For $x = -4$ in $(-\infty, -3)$: $$\frac{(-4 - 2)(-4 - 4)}{-4 + 3} = \frac{(-6)(-8)}{-1} = \frac{48}{-1} = -48 < 0$$ (True) - For $x = 0$ in $(-3, 2)$: $$\frac{(0 - 2)(0 - 4)}{0 + 3} = \frac{(-2)(-4)}{3} = \frac{8}{3} > 0$$ (False) - For $x = 3$ in $(2, 4)$: $$\frac{(3 - 2)(3 - 4)}{3 + 3} = \frac{(1)(-1)}{6} = \frac{-1}{6} < 0$$ (True) - For $x = 5$ in $(4, \infty)$: $$\frac{(5 - 2)(5 - 4)}{5 + 3} = \frac{(3)(1)}{8} = \frac{3}{8} > 0$$ (False) 7. **Check domain restrictions:** $x \neq -3$ because denominator cannot be zero. 8. **Write solution:** The inequality holds for intervals where the expression is negative: $$(-\infty, -3) \cup (2, 4)$$ 9. **Final answer:** $$\boxed{(-\infty, -3) \cup (2, 4)}$$ This is the solution set for the inequality.