The expression $16x^{2}y^{4} - 9$ can be factored as a difference of squares. Let's work through it:

1. Recognize that $16x^{2}y^{4}$ and $9$ are both perfect squares: $16x^{2}y^{4} = (4xy^{2})^2 \quad \text{and} \quad 9 = 3^2$

2. Apply the difference of squares formula: $a^2 - b^2 = (a - b)(a + b)$ Here, $a = 4xy^{2}$ and $b = 3$, so: $16x^{2}y^{4} - 9 = (4xy^{2} - 3)(4xy^{2} + 3)$

Thus, the factored form of $16x^{2}y^{4} - 9$ is: $(4xy^{2} - 3)(4xy^{2} + 3)$

Would you like a detailed explanation of any part of this?

Here are five related questions:

1. How would you factor $25x^{4} - 36y^{4}$?
2. Can you apply the difference of squares to $x^2 - 49$?
3. What is the expanded form of $(a - b)(a + b)$?
4. How can you factor $81x^{6} - 16y^{2}$?
5. Can all quadratic expressions be factored?

Tip: The difference of squares only applies when both terms are perfect squares and there's a subtraction between them.