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Find the zeros: Includes a cube

  1. Sep 20, 2007 #1

    Bo_

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    1. The problem statement, all variables and given/known data

    Find the zeros of the function algebraically.



    2. Relevant equations

    f(x) = 4x^3 - 24x^2 - x + 6


    3. The attempt at a solution

    If all quantities had an x in them, I'd just factor out and x, and treat it as a quadratic. But that freaking 6 is ruining my plan and I'm stuck.
     
    Last edited: Sep 20, 2007
  2. jcsd
  3. Sep 20, 2007 #2

    Dick

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    You'll have to try and factor it. Look for a rational root.
     
  4. Sep 20, 2007 #3

    Bo_

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    k, so I factored it to (1 - 4x^2)(-x - 6)

    ...what are the zeros? 6? Someone help me interpret that...pleeease.
     
  5. Sep 20, 2007 #4

    Dick

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    How did you get the (-x-6) factor? I get just (x-6). Or did you mean to write (-x+6)? Then to find the roots, just set each factor equal to zero, right? The product of things can only be zero if one of the things is zero.
     
  6. Sep 20, 2007 #5

    Bo_

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    Damnit, yeah I meant to have (x - 6).
     
  7. Sep 20, 2007 #6

    Bo_

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    Okay so they're 6, +1/2, -1/2. Correct if I'm wrong. If not, thanks.
     
  8. Sep 20, 2007 #7

    Dick

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    They are correct. Didn't want to just leave you hanging. You can also check for yourself, just put those values into the polynomial and see if you get zero.
     
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