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
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
