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Is there formula for zeres of a cubic polynomial

  1. Oct 18, 2012 #1
    is there any general formula to find out zeros of a cubic polynomial that will give you all the zeros ? if not please tell me what are the different methods to find out the zeros , guessing and trial and error , numerical etc. i want to see where are each methods useful and is there any subcategory of cubic polynomial for which we have some specific method to find out all the zeros ?
     
  2. jcsd
  3. Oct 18, 2012 #2
    Yes. It's long and inefficient. A google search for "cubic root formula" should do the trick; or just look up cubic polynomials on wikipedia.
     
  4. Oct 18, 2012 #3

    arildno

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  5. Oct 18, 2012 #4

    HallsofIvy

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    Just because I enjoy doing it, here is a derivation of Cardano's formula:

    If a and b are any two real numbers then [itex](a- b)^3= a^3- 3a^2b+ 3ab^2- b^3[/itex] and [itex]3ab(a- b)= 3a^2b- 3ab^2[/itex] so that [itex](a- b)^3+ 3ab(a- b)= a^3- b^3[/itex]. If we let x= a- b, m= 3ab, and [itex]n= a^3- b^3[/itex], we have the [itex]x^3- mx= n[/itex]. That is a "reduced" cubic equation. (Reduced because there is no "[itex]x^2[/itex]" term. Given any cubic equation, [itex]x^3+ px^2+ qx+ r= 0[/itex], we can always replace x by y- a, then choose a so that the coefficient of [itex]y^2[/itex] is 0.)

    Now, the question is, suppose we know m and n, can we solve for a and b and so find x? The answer is, of course, yes. From m= 3ab, b= m/3a so that [itex]n= a^3- b^3= a^3- m^3/3^3a^3[/itex] and, multiplying through by [itex]a^3[/itex], [itex]na^3= n(a^3)^2- (m/3)^3= 0.[/itex] That is a quadratic equation for [itex]a^3[/itex] that we can solve using the quadratic equation:
    [tex]a^3= \frac{n\pm\sqrt{n^2- 4(m/3)^3}}{2}= \frac{n}{2}\pm\sqrt{\left(\frac{n}{2}\right)^2- \left(\frac{m}{3}\right)^3}[/tex]

    Since [itex]a^3- b^3= n[/itex],
    [tex]b^3= a^3- n= -\frac{n}{2}\pm\sqrt{\left(\frac{n}{2}\right)^2- \left(\frac{m}{3}\right)^3}[/tex]
     
    Last edited by a moderator: Oct 18, 2012
  6. Oct 26, 2012 #5
    i also enjoyed it :smile: but find little tough to get it ,i need to apply it for a polynomial thanks .
     
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