Question

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

A: 81y^4-16x^2, a perfect square trinomial

B: 81y^4-16x^2, the difference of squares

C: 81y^4-72xy^2-16x^2, a perfect square trinomial

D: 81y^4-72xy^2-16x^2, the difference of squares

Answer 1

Answer: b

Step-by-step explanation: edge 2022

Answer 2

Option B: 81y^4-16x^2, the difference of squares

Explanation:

we know that

The Difference of Squares is two terms that are squared and separated by a subtraction sign

so

`(a+b)(a-b)=(a^(2)-b^(2))`

In this problem we have

`(9y^(2)-4x)(9y^(2)+4x)`

Let

`a=9y^(2)`

`b=4x`

so

`a^(2)=(9y^(2))^(2)=81y^(4)`

`b^(2)=(4x)^(2)=16x^(2)`

substitute

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