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Homework Help: How to factor x^3 - 4x + 4

  1. Dec 8, 2008 #1
    1. The problem statement, all variables and given/known data
    [tex]x^3-4x+4[/tex]


    2. Relevant equations
    None


    3. The attempt at a solution
    I tried using the quadratic formula but it doesnt work. I started with this, but I don't know what to do now.[tex]x^{3}+0x^2-4x+4[/tex]
     
  2. jcsd
  3. Dec 8, 2008 #2

    gabbagabbahey

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    Re: Factoring

    What exactly are you trying to do?!

    [itex]x^3-4x+4[/itex] is an expression, not a problem statement! Do you mean that you are trying to find the roots of the equation [itex]x^3-4x+4=0[/itex]?
     
  4. Dec 8, 2008 #3
    Re: Factoring

    yes I am trying to find the roots of the equation
     
  5. Dec 8, 2008 #4

    gabbagabbahey

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    Re: Factoring

    Well, there are no simple solutions to this cubic equation. You can try using the cubic formula found here.

    Is this the entire question, or is this part of a larger problem? Are you sure you have have written the equation correctly?
     
  6. Dec 8, 2008 #5
    Re: Factoring

    This is the correct equation, and it is the entire question.
     
  7. Dec 8, 2008 #6

    gabbagabbahey

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    Re: Factoring

    Are you asked to use any specific method? (such as approximating the real root using newton's method)
    Are you allowed to use a calculator/computer?
     
  8. Dec 8, 2008 #7
    Re: Factoring

    No specific method is being asked but you can use a simple scientific calculator.
     
  9. Dec 8, 2008 #8

    gabbagabbahey

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    Re: Factoring

    What methods have you been taught in whichever course this is for? Use one of those.
     
  10. Dec 8, 2008 #9
    Re: Factoring

    Edit~
     
  11. Dec 8, 2008 #10

    gabbagabbahey

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    Re: Factoring

    There are no rational roots to this equation, so that method won't work.
     
  12. Dec 9, 2008 #11
    Re: Factoring

    Find the roots using Cardano's method.

    [tex]D=(\frac{4}{2})^2 + (\frac{-4}{3})^3=4 - \frac{64}{27}=\frac{108-64}{27}=\frac{44}{27} > 0[/tex], so it got one rational and two complex roots.

    The rational one is ~(-2.38)
     
  13. Dec 9, 2008 #12

    gabbagabbahey

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    Re: Factoring

    The root (-2.38) is real not rational. The rational roots theorem tells you that the only possible candidates for rational roots of this equation are [itex]\pm 1[/itex], [itex]\pm 2[/itex], and [itex]\pm 4[/itex]....and since none of these are roots; there are no rational roots.
     
  14. Dec 10, 2008 #13
    Re: Factoring

    Rational numbers are also real numbers. So I suppose -2.38....(infinite number of decimals) is irrational number (also real number).
     
  15. Dec 10, 2008 #14

    gabbagabbahey

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    Re: Factoring

    A rational number is a real number that can be expressed as a ratio of two integers....for example, 1/3, 0.25, and 3658793659/548709568017097 are rational numbers.

    However, things like [itex]\pi[/itex], [itex]\sqrt{2}[/itex] and any non-repeating non-terminating decimal are irrational numbers.
     
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