1. **State the problem:** Factor the quadratic expression $n^2 + 6n + 8$. 2. **Recall the factoring formula:** For a quadratic $ax^2 + bx + c$, we look for two numbers that multiply to $ac$ and add to $b$. 3. Here, $a=1$, $b=6$, and $c=8$. We need two numbers that multiply to $1 \times 8 = 8$ and add to $6$. 4. The numbers $2$ and $4$ satisfy this because $2 \times 4 = 8$ and $2 + 4 = 6$. 5. Rewrite the middle term using these numbers: $$n^2 + 2n + 4n + 8$$ 6. Group terms: $$(n^2 + 2n) + (4n + 8)$$ 7. Factor each group: $$n(n + 2) + 4(n + 2)$$ 8. Factor out the common binomial: $$(n + 2)(n + 4)$$ **Final answer:** The factored form of $n^2 + 6n + 8$ is $$\boxed{(n + 2)(n + 4)}$$.