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Math Help - Adding Rational Expressions

  1. #1
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    Adding Rational Expressions

    Problem: \frac{x - 8}{x^2 - 4} + \frac{3}{x^2-2x}

    Answer: \frac{x-3}{x(x+2)}, x\not=2

    What I ended up with: \frac{x^2-5x+6}{x(x^2 - 4)}

    Any help on how to do problems like this would be great as well. I feel like I was doing everything correctly, but now that I'm at the end I don't see any way to make my answer look like the one in the book.
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  2. #2
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    Re: Adding Rational Expressions

    Very close. All you need to do is observe that x^2- 5x+ 6= (x- 2)(x- 3).

    Even if you are not good at factoring in general, you should recognize that x^2- 4= (x- 2)(x+ 2), which you apparently did to get the "least common denominator" and then check for common factors in the numerator and denominator. And you can do that by evaluating 0^2- 5(0)+ 6= 6 so x- 0= x is not a factor. (-2)^2- 5(-2)+ 3= 17 so x+ 2 is not a factor. But 2^2- 5(2)+6= 0 so x- 2 is a factor of x^2- 5x+ 6.
    Thanks from fogownz
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  3. #3
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    Re: Adding Rational Expressions

    Quote Originally Posted by fogownz View Post
    Problem: \frac{x - 8}{x^2 - 4} + \frac{3}{x^2-2x}

    Answer: \frac{x-3}{x(x+2)}, x\not=2

    What I ended up with: \frac{x^2-5x+6}{x(x^2 - 4)}

    Any help on how to do problems like this would be great as well. I feel like I was doing everything correctly, but now that I'm at the end I don't see any way to make my answer look like the one in the book.
    Notice that \displaystyle \begin{align*} x^2 - 4 = (x - 2)(x + 2) \end{align*} and \displaystyle \begin{align*} x^2 - 2x = x(x - 2) \end{align*}. So the lowest common denominator is \displaystyle \begin{align*} x(x-2)(x+2) \end{align*}. You should get into the habit of keeping things factorised until the end, as it makes cancelling a lot easier.
    Thanks from fogownz
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