How to expand to (x squared + 2y) (–2y + x squared)?

= (x² + 2y)(- 2y + x)
= - 2x²y + x³ - 4y² + 2xy
= x³ - 2x²y + 2xy - 4y²
= x²(x - 2y) + 2y(x - 2y)
= (x² - 2y)(x² + 2y)
= x⁴ - 4y²

Answer: x⁴ - 4y²

x^4 + 4x^2y + 4y^2

(x^2 + 2y)(-2y + x^2)
= x^2*-2y - 2y*2y + x^2*x^2 + 2y*x^2
= -2x^2y - 4y^2 + x^4 + 2x^2y
= x^4 - 4y^2

use formula( a^2- b^2) = (a +b)(a-b)
so ans is (x^2)^2 -(2y)^2

(x ^ 2 + 2y) (-2y + x ^ 2)
This can be written as :

(x ^ 2 + 2y) (x ^ 2 - 2y)

Now, you can either

A - Multiply

= x ^ 2 (x ^ 2 - 2y) + 2y (x ^ 2 - 2y)
= x ^ 4 - 2x ^ 2 y + 2x ^ 2 y - 4y ^ 2

-2x ^ 2 y and + 2x ^ 2 y cancel/
= x ^ 4 - 4y ^ 2

B - Use an algebraic identity

Using identity (a + b) (a - b) = a ^ 2 - b ^ 2
(x ^ 2 + 2y) (x ^ 2 - 2y)

a ---> x ^ 2
b ---> 2y

= (x ^ 2) ^ 2 - (2y) ^ 2
= x ^ 4 - 2y ^ 2

Either way, you'll get the same answer.

Hope I helped!

( x ² + 2 y ) ( x ² - 2 y )

x^4 - 4 y^2

(x squared + 2y) (–2y + x squared)
= (x^2 + 2y) (-2y + x^2)
= (x^2 + 2y) (x^2 - 2y)
= (x^2)^2 -(2y)^2
= x^4 - 4y^2

There is a simple equation for this, called the product of a sum and a difference.

It is: (A + B)(A - B) = A² - B²

So in your case, A = x², and B = 2y (note that you need to reverse the terms of the second parenthesis.)

So we end up with (x²)² - (2y)² or x^4 - 4y²

= ( - 2x^2y + x^4 - 4y^2 + 2x^2y)
= x^4 - 4y^2

^ means to the power of =]

(x^2 + 2y) (–2y + x^2)

Re-order:
(x^2 + 2y)(x^2 - 2y)

Use FOIL (First Outer Inner Last):
(x^2)(x^2) + (x^2)(-2y) + (2y)(x^2) + (2y)(-2y)

= x^4 -2x^2y + 2x^2y - 4y^2

= x^4 - 4y^2


This is also known as the difference of two squares.
(x^2 + 2y)(x^2 - 2y)

The first terms multiplied together, minus the last terms multiplied together (the middle terms cancel eachother out):
(x^2)(x^2) - (2y)(2y)

= x^4 - 4y^2