To factor the expression $81y^4 - 16$ using integer coefficients, we again recognize it as a difference of squares.

Step 1: Identify the square terms

• $81y^4$ can be written as $(9y^2)^2$.
• $16$ can be written as $4^2$.

So, the expression can be rewritten as:

$81y^4 - 16 = (9y^2)^2 - 4^2$

Step 2: Apply the difference of squares formula

Using the difference of squares formula $a^2 - b^2 = (a - b)(a + b)$:

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

Step 3: Factor further if possible

Now, let's check if $9y^2 - 4$ and $9y^2 + 4$ can be factored further.

• $9y^2 - 4$ is also a difference of squares because $9y^2 = (3y)^2$ and $4 = 2^2$:

$9y^2 - 4 = (3y - 2)(3y + 2)$

• $9y^2 + 4$ is a sum of squares, which cannot be factored further using real numbers.

Final Factored Form

Thus, the fully factored form of $81y^4 - 16$ is:

$\boxed{(3y - 2)(3y + 2)(9y^2 + 4)}$

Would you like further details or have any questions?

Here are some related questions:

1. How would you factor $25y^4 - 36$?
2. What is the factored form of $81y^4 + 16$?
3. Can you factor $49y^4 - 81$ using a similar approach?
4. How would you factor $64y^4 - 1$?
5. What is the significance of recognizing difference of squares in factoring?

Tip: Always consider checking each factor to see if it can be factored further, especially when dealing with expressions involving higher powers.