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by **bbsa3412** on Sat Mar 28, 2009 10:28 pm

Help me finish this problem....Cant factor by grouping, what other method?

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by **stapel_eliz** on Sun Mar 29, 2009 1:01 am

It looks like the image displays the following:

$\frac{\frac{x\, -\, 2}{x\, +\, 2}\, +\, \frac{x\, -\, 1}{x\, +\, 1}}{\frac{x}{x\, +\, 1}\, -\, \frac{2x\, -\, 3}{x}}$

To simplify, you appear to be multiplying, top and bottom, by the LCM of the various denominators, being:

$x(x\, +\, 2)(x\, +\, 1)$

After multiplying, you get:

$\frac{(x^3\, -\, x^2\, -\, 2x)\, +\, (x^3\, +\, x^2\, -\, 2x)}{(x^3\, +\, 2x^2)\, -\, (2x^3\, +\, 3x^2\, -\, 5x\, -\, 6)}$

So I get the same products that you did. Then:

$\frac{2x^3\, -\, 4x}{-x^3\, -\, x^2\, +\, 5x\, +\, 6}$

Each of the numerator and denominator can be factored, but nothing cancels. You can factor the numerator easily, as you have already displayed. You'll need to use the Rational Roots Test to factor the denominator, and then the remaining quadratic is prime.

$\frac{2x(x\, -\, 2)}{-1(x\, +\, 2)(x^2\, -\, x\, -\, 3)}$

So the above is as "simplified" as you can get. :wink:

by **bbsa3412** on Sun Mar 29, 2009 2:07 am

Thanks for the quick reply!!! I think the reason I was confused with factoring the denominator is because my book does not refer to the rational roots test in prior sections, so I would never had known how to factor it anyway. Thus far it has only demonstrated factoring by common monomial, special products, grouping, and the AC method.

by **stapel_eliz** on Sun Mar 29, 2009 1:12 pm

Long story short: While you should expect to need to factor and simplify this sort of expression on your next test (it's a common "trick"), it is a fact that most of these expressions do not in fact simplify further. Don't be surprised when most of them don't! :wink:

P.S. Thank you for showing your work so nicely!!!

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