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Math Help - How do you find the bounds of real zeros?

  1. #1
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    How do you find the bounds of real zeros?

    the problem is:

    x^3 + x^2 - 32x - 60

    It says something about guessing and checking, but how do I know if the number I pick works?

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  2. #2
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    Divide your cubic by x+3. If it is a zero, it'll reduce to a quadratic.
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  3. #3
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    Quote Originally Posted by galactus View Post
    Divide your cubic by x+3. If it is a zero, it'll reduce to a quadratic.
    How did you get that, though?

    And by cubic, do you mean the whole polynomial?

    edit: I need the upper bound and lower bound.
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  4. #4
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    By the rational root theorem, -60 divides -3. It also divides others. That is where your guess and check comes in. Cubic means it is a third degree polynomial.
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    Quote Originally Posted by galactus View Post
    By the rational root theorem, -60 divides -3. It also divides others. That is where your guess and check comes in. Cubic means it is a third degree polynomial.
    Ah, that makes sense now.

    Thanks!
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    Quote Originally Posted by mankvill View Post
    the problem is:

    x^3 + x^2 - 32x - 60

    It says something about guessing and checking, but how do I know if the number I pick works?

    you can try using the rational roots theorem to check for rational zeros. the roots here might not be rational since they ask for bounds rather than the actual zeros, but who knows. start by plugging in the factors (+/-) of 60 to see if you get zero when you evaluate it. if you do, great, if not, try picking numbers close to that that get you as close to zero as possible from both sides, that will get you the bounds. if you actually find a rational zero, call it a, then you can divide the polynomial by (x - a) and it will reduce it to a quadratic where you can find the other zeros.
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  7. #7
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    blech. could someone walk me through this one?

    x^3 - 5x^2 - 2x + 24
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  8. #8
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    Quote Originally Posted by mankvill View Post
    blech. could someone walk me through this one?

    x^3 - 5x^2 - 2x + 24
    Because this is monic the rational root theorem tells us that if this has a rational root it is in fact an integer and a factor of 24.

    The factors of 24 are \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12, \pm 24 so you try x=\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12, \pm 24 in turn untill you find one that works.

    Once you have found a root r, you know that (x-r) is a factor, and so you can divide the cubic through by the factor to find quadratic x^2+ax+b such that:

    x^3 - 5x^2 - 2x + 24=(x-r)(x^2+ax+b)

    and carry on from there

    RonL
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    Quote Originally Posted by CaptainBlack View Post
    Because this is monic the rational root theorem tells us that if this has a rational root it is in fact an integer and a factor of 24.

    The factors of 24 are \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12, \pm 24 so you try x=\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12, \pm 24these in turn untill you find one that works.

    RonL
    and I do that using synthetic division?
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  10. #10
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    Quote Originally Posted by mankvill View Post
    and I do that using synthetic division?
    yes, or long division or inspection

    RonL
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    okay, thanks a ton!

    one last question and then i'll shut up, i promise.

    I know i'm looking for one that has a remainder of 0, but what about the other one?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by mankvill View Post
    okay, thanks a ton!

    one last question and then i'll shut up, i promise.

    I know i'm looking for one that has a remainder of 0, but what about the other one?
    i don't think i understand your question. all zeros will give a remainder of zero when you plug them in, or divide by the factor we pointed out. if you find a zero, you can reduce it to a quadratic in which case you can find the other zeros by factoring or using the quadratic formula. otherwise, try the guess and check approach i described
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  13. #13
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    Hmm. Okay. I think I might be misunderstanding something.

    Anyway: I plugged in 3 and found out that it works. Do I need to do anymore, or just use that one?
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  14. #14
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    You should now have a quadratic. Solve it by the quadratic formula or whatever and you're done.
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  15. #15
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    according to the post on the previous page, I now have:

    (x - 3)(x^2 + ax + b)

    I'm just so confused.
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