which is equivalent to $(9y^{2}-4x)(9y^{2}+4x)$, and what type of special product is it?\n$81y^{4}-16x^{2}$…

which is equivalent to $(9y^{2}-4x)(9y^{2}+4x)$, and what type of special product is it?\n$81y^{4}-16x^{2}$, a perfect square trinomial\n$81y^{4}-16x^{2}$, the difference of squares\n$81y^{4}-72xy^{2}-16x^{2}$, a perfect square trinomial\n$81y^{4}-72xy^{2}-16x^{2}$, the difference of squares

which is equivalent to $(9y^{2}-4x)(9y^{2}+4x)$, and what type of special product is it?\n$81y^{4}-16x^{2}$, a perfect square trinomial\n$81y^{4}-16x^{2}$, the difference of squares\n$81y^{4}-72xy^{2}-16x^{2}$, a perfect square trinomial\n$81y^{4}-72xy^{2}-16x^{2}$, the difference of squares

Answer

Explanation:

Step1: Apply the difference - of - squares formula

The formula for the difference of squares is ((a - b)(a + b)=a^{2}-b^{2}). Here, (a = 9y^{2}) and (b = 4x). [ \begin{align*} (9y^{2}-4x)(9y^{2}+4x)&=(9y^{2})^{2}-(4x)^{2}\ \end{align*} ]

Step2: Simplify the exponents

Using the power of a power rule ((m^{n})^{p}=m^{np}), we have ((9y^{2})^{2}=9^{2}\times(y^{2})^{2}=81y^{4}) and ((4x)^{2}=16x^{2}) [ (9y^{2})^{2}-(4x)^{2}=81y^{4}-16x^{2} ]

Answer:

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