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How can I find the zeros to this cubic equation?

  1. Feb 18, 2014 #1
    4x^3 + 8x^2 + 41x + 37
     
  2. jcsd
  3. Feb 18, 2014 #2
    Synthetic division looks like a good bet. Although you'd have to try a lot of different roots.
     
  4. Feb 18, 2014 #3

    maajdl

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  5. Feb 18, 2014 #4

    Integral

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    Graph it.
     
  6. Feb 18, 2014 #5

    Mentallic

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    Not really, 37 is prime, so we have the possible rational roots

    [tex]\pm\frac{1,37}{1,2,4}[/tex]

    but since all the coefficients of the cubic are positive, we can toss out the positive roots as possible solutions so all we're left with are 6 possibilities.
     
  7. Feb 18, 2014 #6

    Borek

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  8. Feb 18, 2014 #7
    for any polynomial there's the Horner's method, which is basically trial and error.
    [tex]ax^3 + bx^2 + cx + d = 0[/tex]

    I prefer substituting to remove the squared member from the equation such that
    [tex]y = x + \frac{b}{3a}[/tex] ( Tschirnhausen's substitution I believe)and you would end up with[tex]y^3 + py + q = 0[/tex] then apply something called the Cardano's method or solution, I'm not even sure what it's called in English tbh.

    [tex]x_{1,2,3} = u - \frac{p}{3u} \Longrightarrow u^3 = - \frac{q}{2} + \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}[/tex]fromt he back of my head, this is the simplest way to tackle any cubed polynomial.
     
    Last edited: Feb 18, 2014
  9. Feb 18, 2014 #8

    Mentallic

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    lendav_rott, you should always check for rational roots first because if they exist, then the problem becomes much simpler.
     
  10. Feb 18, 2014 #9
    Ok, yes, you can find the possible candidates for rational roots and test them with the Horner's method, for example, but it's such a tedious process :( Cubics are always tedious, certainly, it will become easier once you determine there are are rational roots to the problem. I guess to each their own. Mathematica or mathcad doesn't complain if I make them do something the long way , might take a few microseconds longer :(
     
    Last edited: Feb 18, 2014
  11. Feb 18, 2014 #10

    ehild

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    It does not hurt to try x=1 or x=-1 before starting some complicated process. Here the root can be only negative, and -1 works !!!!

    ehild
     
  12. Feb 18, 2014 #11
    If, once you find a rational root like -1, you can long divide (x- (root) ) into the original cubic polynomial to reduce it to a quadratic and then solve that by using the quadratic formula in order to find the other two roots.
     
  13. Feb 18, 2014 #12

    HallsofIvy

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    There are, unfortunately, no rational zeros of this polynomial so you will need to use Cardano's formula.
     
  14. Feb 18, 2014 #13

    ehild

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    X=-1: 4x^3 + 8x^2 + 41x + 37 =0 ---> -4+8-41+37 = 4-4 = 0:tongue:

    ehild
     
  15. Feb 18, 2014 #14

    D H

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    Um, no. It's a trivial matter to determine that -1 is a solution: -4+8-41+37 is zero. Or, as ehild put it earlier,
     
  16. Feb 18, 2014 #15

    Ray Vickson

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    Likewise, after I press 'enter', Maple returns the three solutions before my hand can reach my coffee cup. My excuse is that I have been doing this stuff for more than 50 years and have just gotten lazy in my old age. However, I am conflicted about the suggestion of an immediate jump to the use of a CAS by somebody who is just starting out in learning the material.
     
    Last edited: Feb 18, 2014
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