To solve the expression \((2f + 1)(2f - 1)\), we can apply the difference of squares formula. The difference of squares states that:

\[
(a + b)(a - b) = a^2 - b^2
\]

Here, \(a = 2f\) and \(b = 1\). Plugging these into the formula:

\[
(2f + 1)(2f - 1) = (2f)^2 - 1^2
\]

Now, calculate each square:

\[
(2f)^2 = 4f^2 \quad \text{and} \quad 1^2 = 1
\]

So the expression simplifies to:

\[
4f^2 - 1
\]

This is the simplified form of the expression.

Would you like more details or have any questions about this process?

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Here are 5 related questions you might find interesting:

1. What is the difference of squares formula, and how is it derived?
2. How would the expression change if it were \((2f + 2)(2f - 2)\)?
3. What is the general approach to expanding binomials?
4. How would you solve for \(f\) if the expression \((2f + 1)(2f - 1) = 0\)?
5. Can the difference of squares be factored further if more complex numbers or variables are involved?

**Tip:** When dealing with expressions that involve binomials, always check if the difference of squares formula can be applied for quick simplification!

Understand how to simplify the algebraic expression (2f+1)(2f-1) using the difference of squares formula. This problem explores key algebraic concepts and is suitable for students in Grades 8-10.

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