Rational expression?

Question

I need have tried to solve this for 2 days....I cannot get it to come out correctly. Can someone please help???

2x^2-2y^2 divided by x^2+2xy+y^2
_______ _________
14x^2y^4 35xy^3

I dont know how to put a fraction in.....but the line in between is the fraction line. I have factored both numerators before I flipped the second one...and there I am stuck!

Answer 1

Factor after you've flipped the denominator and simplified:
(2x^2 - 2y^2)(35xy^3)/(x^2 + 2xy + y^2)(14x^2y^4) =
(2x^2 - 2y^2)(5)/(x^2 + 2xy + y^2)(2xy) =
2(x + y)(x - y)(5)/(x + y)^2(2xy)
5(x - y)/(xy)(x + y)

Answer 2

(If I am reading this right),

I think you are on the right track.

(2x^2 - 2y^2) / (14x^2y^4) divided by (x^2 + 2xy + y^2) / 35xy^3
Just because this is difficult to write on the computer, let's flip the second one first.
(2x^2 - 2y^2)(35xy^3) / (14x^2y^4)(x^2 + 2xy + y^2)

Now let's factor those like you already did:
2(x^2 - y^2)(35xy^3) / 14x^2y^4(x+y)^2
2(x+y)(x-y)(35xy^3) / 14x^2y^4(x+y)^2
Cancel (x+y)
2(x-y)(35xy^3) / 14x^2y^4(x+y)
Cancel x
2(x-y)(35y^3) / 14xy^4(x+y)
Cancel y^3
2(x-y)(35) / 14xy(x+y)
2(x-y)(7*5) / (2*7)xy(x+y)
Cancel the 2 and the 7
(x-y)(5) / xy(x+y)

Good luck!

Answer 3

First fraction:
2(x+y)(x-y)/14(x^2)(y^4) = (x+y)(x-y)/[7(x^2)(y^4)]

Second fraction:
(x+y)^2/[35xy^3]

Now dividing ==> "flipping" the second fraction we get
(x+y)(x-y)(35xy^3)/[7(x^2)(y^4)][(x+y)^2]
= 5(x-y)/[xy(x+y)]

It sure looks right to me.