1. **State the problem:** We are asked to factor the polynomial $49n^2 + 14n + 1$. 2. **Recall the formula and rules:** For a quadratic polynomial $ax^2 + bx + c$, factoring involves finding two binomials $(pn + q)(rn + s)$ such that: $$pr = a, \quad qs = c, \quad ps + qr = b$$ 3. **Apply to the problem:** Here, $a=49$, $b=14$, and $c=1$. 4. **Find factors of $a$ and $c$:** - Factors of $49$ are $7$ and $7$. - Factors of $1$ are $1$ and $1$. 5. **Try binomials:** $$(7n + 1)(7n + 1)$$ 6. **Check by expansion:** $$7n \times 7n = 49n^2$$ $$7n \times 1 = 7n$$ $$1 \times 7n = 7n$$ $$1 \times 1 = 1$$ Sum of middle terms: $7n + 7n = 14n$ 7. **Conclusion:** The factorization is: $$49n^2 + 14n + 1 = (7n + 1)^2$$