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We are tasked with factoring the quadratic expression \( 49a^2 - 14a + 1 \).

To factor this, we can look for two binomials of the form:
\[
(7a + m)(7a + n)
\]
When expanded, this will give:
\[
7a \cdot 7a = 49a^2
\]
\[
7a \cdot n + 7a \cdot m = 7a(n + m)
\]
\[
m \cdot n = 1
\]
We need \( n + m = -14 \) (the coefficient of \( a \)) and \( mn = 1 \) (the constant term).

Now, solving for the factors:
- We need two numbers that multiply to 1 and add to -14. The numbers \( -7 \) and \( -7 \) satisfy this because:
\[
-7 \cdot -7 = 49
\]
\[
-7 + -7 = -14
\]

Thus, the factorization is:
\[
49a^2 - 14a + 1 = (7a - 1)(7a - 1) = (7a - 1)^2
\]

So, the factored form is:
\[
(7a - 1)^2
\]

Let me know if you'd like further explanation on any step.

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Here are some related questions you might want to consider:
1. How do we know if a quadratic expression can be factored easily?
2. What if the quadratic expression was not factorable using integers? What would we do next?
3. How would we check that our factorization is correct?
4. Can all quadratic expressions be factored into real numbers?
5. How does completing the square relate to factoring?

**Tip:** Always double-check your factorization by expanding it back out to the original expression!

Learn how to factor the quadratic expression 49a^2 - 14a + 1. This problem demonstrates key algebraic techniques, including factoring perfect square trinomials. Ideal for students in Grades 9-10.

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