SOLUTION: b^2/b^2-4 - 4/b^2-2b I know you have to find the common denominators by factoring the denominators so it ends up being: b^2/(b-2)(b+2) - 4/(b-2)(b+b) but after that step, I don't

Question 127609This question is from textbook College Algebra
: b^2/b^2-4 - 4/b^2-2b
I know you have to find the common denominators by factoring the denominators so it ends up being:
b^2/(b-2)(b+2) - 4/(b-2)(b+b) but after that step, I don't know what to do and which numbers to pick to get the same denominator. This question is from textbook College Algebra

Answer by oscargut(2103) (Show Source): You can put this solution on YOUR website!

First of all you have a mistake
b^2/b^2-4 - 4/b^2-2b =b^2/(b-2)(b+2) - 4/(b-2)(b)
the denominators are (b-2)(b+2) and (b-2)(b) to find the common denominator
you have to select (b-2) (because is in both denominators) and (b+2) and (b)
so the common denominator is (b-2)(b+2)b
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