1. **State the problem:** Simplify the expression $$\frac{x^4 + 8x^3 + 24x^2 + 32x + 16}{x^3 + 6x^2 + 12x + 8}$$ and determine its domain. 2. **Recognize the polynomials:** Both numerator and denominator look like expansions of binomials raised to powers. 3. **Factor the numerator:** Notice that $$x^4 + 8x^3 + 24x^2 + 32x + 16$$ matches the expansion of $$(x+2)^4$$ because: $$ (x+2)^4 = x^4 + 4 \cdot x^3 \cdot 2 + 6 \cdot x^2 \cdot 2^2 + 4 \cdot x \cdot 2^3 + 2^4 = x^4 + 8x^3 + 24x^2 + 32x + 16 $$ 4. **Factor the denominator:** Similarly, $$x^3 + 6x^2 + 12x + 8$$ matches $$(x+2)^3$$ because: $$ (x+2)^3 = x^3 + 3 \cdot x^2 \cdot 2 + 3 \cdot x \cdot 2^2 + 2^3 = x^3 + 6x^2 + 12x + 8 $$ 5. **Rewrite the fraction:** $$ \frac{(x+2)^4}{(x+2)^3} $$ 6. **Simplify the fraction:** $$ \frac{\cancel{(x+2)^3} \cdot (x+2)}{\cancel{(x+2)^3}} = x+2 $$ 7. **Domain considerations:** The original denominator is zero when $$x+2=0 \Rightarrow x=-2$$, so the expression is undefined at $$x=-2$$. 8. **Final answer:** $$ \frac{x^4 + 8x^3 + 24x^2 + 32x + 16}{x^3 + 6x^2 + 12x + 8} = x+2 \quad \text{for} \quad x \neq -2 $$ **Summary:** The simplified expression is $$x+2$$ with the restriction that $$x \neq -2$$ because the original denominator is zero there.