SOLUTION: ((x^2-4)/(x^2-4x-21))/((x^2-7x+10)/(X^2-2x-35)) This particular problem has been giving me lots of trouble. I figured that you would have to factor the bottom of the fractions

Algebra ->  Proportions  -> Lessons -> SOLUTION: ((x^2-4)/(x^2-4x-21))/((x^2-7x+10)/(X^2-2x-35)) This particular problem has been giving me lots of trouble. I figured that you would have to factor the bottom of the fractions       Log On


   

Question 22839: ((x^2-4)/(x^2-4x-21))/((x^2-7x+10)/(X^2-2x-35))
This particular problem has been giving me lots of trouble. I figured that you would have to factor the bottom of the fractions to make them have four terms but I am stuck there. Id appreciate the help Thanks!
Answer by rapaljer(4671) About Me  (Show Source):
You can put this solution on YOUR website!
Is this a complex fraction that looks like this:
%28%28x%5E2-4%29%2F%28x%5E2-4x-21%29%29%2F%28%28x%5E2-7x%2B10%29%2F%28x%5E2-2x-35%29%29

If so, then notice that you have a numerator which is a fraction and a denominator which is a fraction. Notice also that the middle fraction line separating the numerator and the denominator is the longest line. Now, just "unstack" the problem by writing this in the form of a
(NUMERATOR) divided by (DENOMINATOR).
+%28x%5E2-4%29%2F%28x%5E2-4x-21%29 divided+by +%28x%5E2-7x%2B10%29%2F%28x%5E2-2x-35%29
Now, invert the second fraction and multiply. While you are at it, factor all the trinomials:


Divide out the (x-7) and the (x-2) factors, and you should be left with %28%28x%2B2%29%2A%28x%2B5%29%29%2F%28%28x%2B3%29%2A%28x-5%29%29 for your final answer.

R^2 at SCC