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:

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

Step-by-step 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

$(9y^{2}-4x)(9y^{2}+4x)=81y^{4}-16x^{2}$