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Math Help - Another basic factoring question

  1. #1
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    Another basic factoring question

    If im presented with
    \frac{(x+1)(x-3)}{(x+5)(x+1)} we can reduce that to \frac{x-3}{x+5} canceling out the x+1 from the numerator/ denominator

    however if we were to take this same problem without it being joined by multiplication the above would not be possible e.g.

    \frac{(x+1)+(x-3)}{(x+5)(x+1)} then cancellation is no longer possible

    instead here we have the following after combing like terms in the numerator
    \frac{(x+1)+(x-3)}{(x+5)(x+1)} =\frac{2x-2}{(x+5)(x+1)}

    correct?
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  2. #2
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    Quote Originally Posted by allyourbass2212 View Post
    If im presented with
    \frac{(x+1)(x-3)}{(x+5)(x+1)} we can reduce that to \frac{x-3}{x+5} canceling out the x+1 from the numerator/ denominator

    however if we were to take this same problem without it being joined by multiplication the above would not be possible e.g.

    \frac{(x+1)+(x-3)}{(x+5)(x+1)} then cancellation is no longer possible

    instead here we have the following after combing like terms in the numerator
    \frac{(x+1)+(x-3)}{(x+5)(x+1)} =\frac{2x-2}{(x+5)(x+1)}

    correct?
    yes
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  3. #3
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    Yes but you can factor further....
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