1. **State the problem:** Solve the inequality $$\frac{x^2+5x+6}{x+3} > 0$$. 2. **Factor the numerator:** The quadratic $x^2+5x+6$ factors as $$(x+2)(x+3)$$. 3. **Rewrite the inequality:** $$\frac{(x+2)(x+3)}{x+3} > 0$$ 4. **Simplify the expression:** Since $x \neq -3$ (denominator cannot be zero), we can cancel $x+3$: $$\frac{\cancel{(x+3)}(x+2)}{\cancel{(x+3)}} > 0 \implies x+2 > 0$$ 5. **Solve the simplified inequality:** $$x+2 > 0 \implies x > -2$$ 6. **Consider the domain restriction:** The original denominator $x+3 \neq 0$, so $x \neq -3$. 7. **Analyze the sign around $x=-3$:** The original expression is undefined at $x=-3$, so we exclude it. 8. **Check intervals:** - For $x < -3$, numerator $(x+2)(x+3)$ is positive or negative? - At $x=-4$, numerator: $(-4+2)(-4+3) = (-2)(-1) = 2 > 0$. - Denominator $x+3 = -1 < 0$. - So fraction $>0$? Numerator positive, denominator negative, fraction negative. - For $-3 < x < -2$: - At $x=-2.5$, numerator: $(-2.5+2)(-2.5+3) = (-0.5)(0.5) = -0.25 < 0$. - Denominator $x+3 = 0.5 > 0$. - Fraction negative. - For $x > -2$: - At $x=0$, numerator: $(0+2)(0+3) = 2*3=6 > 0$. - Denominator $0+3=3 > 0$. - Fraction positive. 9. **Combine intervals where fraction > 0:** Only for $x > -2$. 10. **Final solution:** $$(-3, -2) \cup (-2, +\infty) = (-3, +\infty) \setminus \{-3\}$$ Since $x=-3$ is excluded, the solution is: $$(-3, -2) \cup (-2, +\infty)$$ **Answer:** $(-3, -2) \cup (-2, +\infty)$ This matches the choice $(-3, -2) \cup (-2, +\infty)$.