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

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

In this problem we have

Let

so

substitute

d, x=8

step-by-step explanation:

simplifying

4x + -5 = 2x + 11

reorder the terms:

-5 + 4x = 2x + 11

reorder the terms:

-5 + 4x = 11 + 2x

solving

-5 + 4x = 11 + 2x

solving for variable 'x'.

move all terms containing x to the left, all other terms to the right.

add '-2x' to each side of the equation.

-5 + 4x + -2x = 11 + 2x + -2x

combine like terms: 4x + -2x = 2x

-5 + 2x = 11 + 2x + -2x

combine like terms: 2x + -2x = 0

-5 + 2x = 11 + 0

-5 + 2x = 11

add '5' to each side of the equation.

-5 + 5 + 2x = 11 + 5

combine like terms: -5 + 5 = 0

0 + 2x = 11 + 5

2x = 11 + 5

combine like terms: 11 + 5 = 16

2x = 16

divide each side by '2'.

a - base

b - exponent

so,