SOLUTION: (y^-5 - 3y^5x^-2 + y^5) x^-2 ________________________________________ (x^2 - p + x^-2) y^5 On this problem I am supposed to use the distributive p

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Question 66818This question is from textbook Algebra 1
: (y^-5 - 3y^5x^-2 + y^5) x^-2
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(x^2 - p + x^-2) y^5
On this problem I am supposed to use the distributive property and write the answer with all exponents positive. I think I would multiply x^-2 by everything on the top and multiply y^5 by everything on the bottom. Where I get confused is changing all exponents to positive. If they're negative I move them to the opposite side and change the sign, right? Somehow, I'm just not getting the right answer. Thanks.This question is from textbook Algebra 1

Answer by stanbon(57219) (Show Source):
You can put this solution on YOUR website!
(y^-5 - 3y^5x^-2 + y^5) x^-2
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(x^2 - p + x^-2) y^5
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Put that x^-2 in the denominator and multiply thru by it to get:
(1/y^5 - 3y^5/x^2 + y^5)
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(x^4 - x^2p + 1)y^5
There is not much more you can do with this except distribute in the denominator to get:
[(x^2 - 3y^10 + x^2y^10)
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(x^4 -x^2p + 1)x^2y^10
[(x^2 - 3y^10 + x^2y^10)
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[x^6y^10 - x^4py^10 +x^2y^10]

I'm sure you could stop at a different point
but this is as good as any.
It's a mess of a problem.
Cheers,
Stan H.