Question

which is equivalent to $(9y^2 - 4x)(9y^2 + 4x)$, and what type of special product is it?

• $81y^4 - 16x^2$, a perfect square trinomial
• $81y^4 - 16x^2$, the difference of squares
• $81y^4 - 72xy^2 - 16x^2$, a perfect square trinomial
• $81y^4 - 72xy^2 - 16x^2$, the difference of squares

Explanation:

Step1: Recall the difference of squares formula

The difference of squares formula is \((a - b)(a + b)=a^{2}-b^{2}\). In the given expression \((9y^{2}-4x)(9y^{2}+4x)\), we can identify \(a = 9y^{2}\) and \(b = 4x\).

Step2: Apply the difference of squares formula

First, calculate \(a^{2}\): \((9y^{2})^{2}=9^{2}\times(y^{2})^{2}=81y^{4}\).
Then, calculate \(b^{2}\): \((4x)^{2}=4^{2}\times x^{2}=16x^{2}\).
Using the formula \((a - b)(a + b)=a^{2}-b^{2}\), we get \((9y^{2}-4x)(9y^{2}+4x)=(9y^{2})^{2}-(4x)^{2}=81y^{4}-16x^{2}\).
Also, since the form is \((a - b)(a + b)\), it is the difference of squares, not a perfect square trinomial (a perfect square trinomial has the form \((a\pm b)^{2}=a^{2}\pm2ab + b^{2}\), which is a trinomial, but our result is a binomial, so it's the difference of squares).

Answer:

B. \(81y^{4}-16x^{2}\), the difference of squares