The expression you've provided is:

$4x^2 - 81$

This is a difference of squares, which can be factored. Recall that a difference of squares has the general form:

$a^2 - b^2 = (a - b)(a + b)$

In your case, $4x^2$ is a perfect square, and $81$ is also a perfect square:

$4x^2 = (2x)^2 \quad \text{and} \quad 81 = 9^2$

Now we can apply the difference of squares formula:

$4x^2 - 81 = (2x - 9)(2x + 9)$

So, the factored form of the expression is:

$(2x - 9)(2x + 9)$

Would you like further clarification or have any questions? Here are some related questions you may find interesting:

1. What is the solution to $4x^2 - 81 = 0$?
2. How does factoring help solve quadratic equations?
3. What is the general formula for factoring a difference of squares?
4. Can this approach be applied to higher degree polynomials?
5. How can completing the square be used to solve quadratic equations?

Tip: Always look for patterns like the difference of squares or perfect square trinomials when factoring, as they simplify the process!