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

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) (Show Source): You can put this solution on YOUR website!
Is this a complex fraction that looks like this:

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).

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 for your final answer.

R^2 at SCC

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