Question

Which is equivalent to (9 y squared minus 4 x)(9 y squared + 4 x), and what type of special product is it?

Answer 1

Answer:

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

and it is the special factor product that leads to a difference of squares

Step-by-step explanation:

The product: $(9y^2-4x)\,(9y^2+4x)$

is a product of the form:

$(a-b)\,(a+b) = a^2-b^2$

which leads as shown to a difference of squares. So they binomials $(a-b)$ and $(a+b)$  are the factors of the difference of squares $a^2-b^2$.

In our case, the product:

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