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Specific factoring question

  1. Sep 5, 2015 #1
    So given the problem 2(x+3)(x-2)/(x-2)≥0

    How come we cant factor x-2 in the numerator against x-2 in the denominator? I mean theyre all just factors right? Does it have something to do about it being an inequality? We got this problem on a test yesterday, and the correct way to solve it was to leave it as i originally wrote it, and then put it on a number line. Why is it wrong to just factor and solve the inequality from there?
     
  2. jcsd
  3. Sep 5, 2015 #2
    It's because that the inequality ##\frac{2(x+3)(x-2)}{x-2}\ge 0## doesn't have the solution ##x=2## for the denominator shouldn't be ##0.## But, the inequality ##2(x+3)\ge 0## does.
    So, if you want to divide something in numerator and denominator simultaneously, you should confirm it would not matter its original solution, that is, think clearly.
     
  4. Sep 5, 2015 #3

    BvU

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    What you call factoring is in fact division by x-2. This can only be done if x is not equal to 2 (you can not divide by 0, not even when you do it both on top and on the bottom). If you solve x+3 > 0 you get x > -3 and have missed excluding this x=2 for the set of x that satisfies the original inequality.

    [edit] slow typist. Good thing the replies agree.
     
  5. Sep 5, 2015 #4
    Oh i see now. Im aware that you cant divide or multiply by the denominator on both sides of the inequality sign, because you cant know if you have to swap the inequality sign or not. I just wasnt aware that i was doing that here.
     
  6. Sep 5, 2015 #5

    BvU

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    Do I understand this ? you want to change ##2(x+3)(x-2)/(x-2)\ge 0## into ##2(x+3)(x-2) \ge 0 \; (x-2) ## and worry about the ##\ge## ?
     
  7. Sep 5, 2015 #6
    No, i meant that dividing x-2 by x-2 would leave uncertainty about whether or not you have to swap the inequality sign, like you guys said.
     
  8. Sep 5, 2015 #7

    BvU

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    It leaves the inequality sign intact. But you can't do it if x-2 is 0, that's the crux.
     
  9. Sep 5, 2015 #8
    Oh, i was thinking about the fact that if you mulitply or divide by something negative on both sides, you have to swap the inequality sign, but i see now that that is not relevant here because you only divide on one side.
     
  10. Sep 5, 2015 #9

    BvU

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    OK, I'm reassured now. Well done.
     
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