Factor: 4x^2 - 25y^2?

(2x+5y)(2x-5y) Perfect squares

The steps are really simple.
Do a factor tree for each i.e. factors of 4 then factors of 25

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

4x^2 - 25y^2
= 4x^2 + 10xy - 10xy - 25y^2
= (2x + 5y)(2x - 5y)

a² - b² = (a - b) (a + b)
(2x)² - (5y)² = (2x - 5y )(2x + 5y)

(2x)^2-(5x)^2
=(2x+5y)(2x-5y)

To get to the answer, think about how these two parts would go together in a form of (ax - by)(ax + by).
One is plus and the other minus because the middle terms will have to cancel out because you have no middle term in
4x^2 - 25y^2

Now what squared is 4? 2
What squared is 25? 5

So you can see that you have to have a = 2 (because the 4 is with the x) and b = 5 (because the 25 is with the y).

Then the answer is: (2x - 5y)(2x + 5y)

When you multiply this out (using foil), you will get:

4x^2 -10xy + 10xy - 25y^2

See here that the middle terms cancel out, which is what we want.

using the principle x^2 - y^2 = (x-y)(x+y)....u get (2x-5y)(2x+5y)

This is a one-step working:
4x^2 - 25y^2 = (2x+5y)(2x-5y)

You just have to recognise that this is of the form
(a^2 - b^2) = (a+b)(a-b)

These can both be easily square rooted with a change of sign in the middle.

You automatically know by looking at it, there are no "steps".

4x^2 - 25y^2 = (2x)^2 - (5y)^2 = (2x + 5y)(2x - 5y)
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Ideas: Use the formula for the difference of two squares: A^2 - B^2 = (A+B)(A-B)