5y^8-125

*you have to factor completley

easy. Take out the five:
5(y^8-25)

Now, isnt y8 the same as y4 )2 ? That is a square, so you have that y4 squared minus another perfect square? That factor as the difference of two squares. After doing that, you may have another difference of squares to factor.

So would it be 5(y^2-5)

To factor completely, let's continue with your progress:

Starting with 5(y^8-25), as you correctly did, we can factor out the common factor of 5:

5(y^8-25)

Next, you correctly recognized that y^8 can be rewritten as (y^4)^2, which is a perfect square. Therefore, we have:

5((y^4)^2-25)

Now, we have a difference of squares, where (y^4)^2 is a perfect square and 25 is also a perfect square (5^2). To factor a difference of squares, we use the formula a^2 - b^2 = (a + b)(a - b). Applying this formula, we have:

5((y^4 + 5)(y^4 - 5))

Now, we have factored the expression completely as 5(y^4 + 5)(y^4 - 5).