SOLUTION: Someone Please check my answer and tell me what i did wrong and how i am suppose to do it.Thanks!, 1.) (x^2-)^4(2x)-x^2(4)(x^2-1)^3(2x) divided by (x^2-1)8 2x(x^2-1)^3[x^2-1)

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Question 718268: Someone Please check my answer and tell me what i did wrong and how i am suppose to do it.Thanks!,
1.) (x^2-)^4(2x)-x^2(4)(x^2-1)^3(2x) divided by (x^2-1)8
2x(x^2-1)^3[x^2-1)-4x^2} divided by x^2-1)^8
=2x(x^2-1)^3-4x^2 divided by (x^2-1)^7
Answer by jsmallt9(3758) (Show Source):
You can put this solution on YOUR website!
1.) (x^2-1)^4(2x)-x^2(4)(x^2-1)^3(2x) divided by (x^2-1)^8
The first step is very good (because a lot of people have trouble seeing how to factor like this):
2x(x^2-1)^3[(x^2-1)-4x^2}] divided by (x^2-1)^8

But you went wrong with:
2x(x^2-1)^3-4x^2 divided by (x^2-1)^7
Apparently you canceled the (x^2-1) in the brackets with one of the eight (x^2-1)'s in the denominator. The error is that only factors can be canceled and the (x^2-1) you canceled in the numerator is not a factor of the numerator!

What you should do is combine like terms in the bracket:
2x(x^2-1)^3[-3x^2-1] divided by (x^2-1)^8
Then you cancel all three (x^2-1)'s in the numerator (next to the 2x) with three of the eight (x^2-1)'s in the denominator:
2x[-3x^2-1] divided by (x^2-1)^5