1. **State the problem:** Find the horizontal asymptote, removable and non-removable restrictions, and simplified factored form of the rational function $$y = \frac{x^3 + 4x^2 - 4x - 16}{x^2 + 5x + 6}$$ 2. **Factor numerator and denominator:** Numerator: Factor by grouping $$x^3 + 4x^2 - 4x - 16 = (x^3 + 4x^2) - (4x + 16) = x^2(x + 4) - 4(x + 4) = (x + 4)(x^2 - 4)$$ Further factor $x^2 - 4$ as difference of squares: $$x^2 - 4 = (x - 2)(x + 2)$$ So numerator factors to: $$ (x + 4)(x - 2)(x + 2) $$ Denominator: $$x^2 + 5x + 6 = (x + 2)(x + 3)$$ 3. **Simplified factored form:** Cancel common factors between numerator and denominator: Common factor: $(x + 2)$ Write the cancellation step: $$y = \frac{(x + 4)(x - 2)\cancel{(x + 2)}}{\cancel{(x + 2)}(x + 3)}$$ Simplified form: $$y = \frac{(x + 4)(x - 2)}{x + 3}$$ 4. **Restrictions:** - Denominator zeroes are restrictions: $x = -2$ and $x = -3$ - Since $(x + 2)$ was canceled, $x = -2$ is a removable restriction (hole) - $x = -3$ remains in denominator, so it is a non-removable restriction (vertical asymptote) 5. **Horizontal asymptote:** Degree numerator (original): 3 Degree denominator (original): 2 Since degree numerator $>$ degree denominator, horizontal asymptote is $y = DNE$ **Final answers:** Horizontal Asymptote: $y = \text{DNE}$ Removable restriction: $x = -2$ Non-removable restriction: $x = -3$ Simplified factored form: $$y = \frac{(x + 4)(x - 2)}{x + 3}$$