81y²-16 factor it using integer coefficient 

To factor the expression \( 81y^2 - 16 \) using integer coefficients, we can recognize it as a difference of squares. The difference of squares formula is:

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

In this case:

- \( a^2 = 81y^2 \), so \( a = 9y \)
- \( b^2 = 16 \), so \( b = 4 \)

Now, apply the difference of squares formula:

\[
81y^2 - 16 = (9y - 4)(9y + 4)
\]

So, the factored form is:

\[
\boxed{(9y - 4)(9y + 4)}
\]

Would you like further details or have any questions? 

Here are some related questions:
1. How would you factor \( 100y^2 - 25 \) using integer coefficients?
2. Can the difference of squares method be applied to \( 81y^2 + 16 \)?
3. How would you factor \( 49y^2 - 36 \)?
4. What if the expression was \( 81y^2 - 25 \), how would it factor?
5. What are other methods of factoring quadratic expressions besides the difference of squares?

**Tip:** Always check if the expression can be simplified or recognized as a special form like a difference of squares before attempting other factoring methods.

Learn how to factor the expression 81y² - 16 using integer coefficients with a detailed step-by-step solution. This problem involves applying the difference of squares formula in algebra, specifically dealing with quadratic equations. Suitable for high school students.

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