We are asked to factor the polynomial $81x^2 - 16$. Let's follow these steps to solve the problem.

Step 1: Recognize it as a difference of squares.

The given polynomial is in the form $a^2 - b^2$, where:

• $a^2 = (9x)^2$
• $b^2 = 4^2$

So, we can rewrite the expression as: $81x^2 - 16 = (9x)^2 - 4^2$

Step 2: Apply the difference of squares formula.

The difference of squares formula is: $a^2 - b^2 = (a - b)(a + b)$

Substitute $a = 9x$ and $b = 4$: $81x^2 - 16 = (9x - 4)(9x + 4)$

Thus, the factored form is: $81x^2 - 16 = (9x - 4)(9x + 4)$

Final Answer:

$81x^2 - 16 = (9x - 4)(9x + 4)$

Would you like further explanation on this process?

Follow-up Questions:

1. How do you factor more complex polynomials?
2. What is the difference of squares rule, and when can it be applied?
3. How do you determine if a polynomial is factorable?
4. What other types of polynomial factorization exist?
5. Can you apply the difference of squares to cubic or quartic expressions?

Tip:

Always check for common factors before applying any factoring techniques.