1. **State the problem:** Find the limit $$\lim_{x \to -1} \frac{x^3 + 4x^2 + 5x + 2}{2x^2 + 3x + 1}$$. 2. **Check direct substitution:** Substitute $x = -1$ into numerator and denominator. Numerator: $$(-1)^3 + 4(-1)^2 + 5(-1) + 2 = -1 + 4 - 5 + 2 = 0$$ Denominator: $$2(-1)^2 + 3(-1) + 1 = 2 - 3 + 1 = 0$$ Since direct substitution gives $\frac{0}{0}$, an indeterminate form, we need to simplify. 3. **Factor numerator and denominator:** Numerator: $x^3 + 4x^2 + 5x + 2$ Try factoring by grouping: $$x^3 + 4x^2 + 5x + 2 = (x^3 + 4x^2) + (5x + 2) = x^2(x + 4) + 1(5x + 2)$$ This does not factor nicely by grouping, so try rational root theorem. Test $x = -1$: $$(-1)^3 + 4(-1)^2 + 5(-1) + 2 = 0$$ So $(x + 1)$ is a factor. Divide numerator by $(x + 1)$: Using synthetic division: Coefficients: 1 | 4 | 5 | 2 -1 | -1 | -3 | -2 Sum: 1 | 3 | 2 | 0 So numerator factors as: $$ (x + 1)(x^2 + 3x + 2) $$ Further factor $x^2 + 3x + 2$: $$x^2 + 3x + 2 = (x + 1)(x + 2)$$ So numerator: $$ (x + 1)^2 (x + 2) $$ Denominator: $2x^2 + 3x + 1$ Factor: $$2x^2 + 3x + 1 = (2x + 1)(x + 1)$$ 4. **Simplify the fraction:** $$\frac{(x + 1)^2 (x + 2)}{(2x + 1)(x + 1)} = \frac{\cancel{(x + 1)} (x + 1)(x + 2)}{(2x + 1) \cancel{(x + 1)}} = \frac{(x + 1)(x + 2)}{2x + 1}$$ 5. **Evaluate the limit by direct substitution now:** $$\lim_{x \to -1} \frac{(x + 1)(x + 2)}{2x + 1} = \frac{(-1 + 1)(-1 + 2)}{2(-1) + 1} = \frac{0 \times 1}{-2 + 1} = \frac{0}{-1} = 0$$ **Final answer:** $$\boxed{0}$$