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.