Using factorization what is 5 minus 125n squared

We are tasked with factorizing the expression \(5 - 125n^2\). Let's go through the process step by step.

1. First, observe that both terms have a common factor of 5. So, we can factor that out:
   \[
   5 - 125n^2 = 5(1 - 25n^2)
   \]

2. Now, focus on the expression inside the parentheses: \(1 - 25n^2\). This is a difference of squares, which follows the formula:
   \[
   a^2 - b^2 = (a - b)(a + b)
   \]
   Here, \(a = 1\) and \(b = 5n\), because:
   \[
   1 - 25n^2 = (1)^2 - (5n)^2
   \]
   Applying the difference of squares formula:
   \[
   1 - 25n^2 = (1 - 5n)(1 + 5n)
   \]

3. Substitute this back into the original factored expression:
   \[
   5(1 - 25n^2) = 5(1 - 5n)(1 + 5n)
   \]

So, the factorization of \(5 - 125n^2\) is:
\[
5(1 - 5n)(1 + 5n)
\]

Would you like more details on this process, or do you have any questions?

Here are some related questions for further practice:
1. Factor \(4 - 36x^2\) using the difference of squares.
2. Factor \(9a^2 - 16b^2\).
3. What is the factored form of \(16 - 81y^2\)?
4. Factor \(3 - 48p^2\).
5. Factor \(49z^2 - 64\).

**Tip**: Always look for common factors before applying more advanced factorization techniques like the difference of squares.

Learn how to factor the expression 5 - 125n² using the difference of squares and factoring common factors. This problem explains step-by-step how to simplify and factor complex expressions. Suitable for students in Grades 8-10.

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