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Rational zeros theorem fails?

  1. Dec 14, 2011 #1
    When solving 3x^3+4x^2-7x+2 the rational zeros theorem says there can only be a possibility of zeros at plus minus 1, 2, 1/3, and 2/3

    but for some reason a zero appears at about -2.4 (or more accurately -1-squareroot of 2).

    How can this be? It states that if there are any zeros they will be listed by the theorem. It does not state the possibility of zeros lying outside that theorem.
     
  2. jcsd
  3. Dec 14, 2011 #2
    The theorem lists all the possible rational roots. If a root is not rational, it won't be listed.
     
  4. Dec 14, 2011 #3

    D H

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    The rational zeros theorem says that the any rational roots must be of the form±p/q. It doesn't say a thing about irrational roots.
     
  5. Dec 14, 2011 #4
    Then it seems a rather clumsy thing to use for solving polynomials. Is there a better theorem out there for this purpose?
     
  6. Dec 14, 2011 #5
    How is it clumsy? It's very simple and elegant, and it completely describes all possible rational roots of the polynomial. If you want every root of a of a polynomial of degree 2 through 4, then just use the quadratic/cubic/quartic formulas. If you want every solution to a polynomial of degree >4 then too bad, a general solution doesn't exist.
     
  7. Dec 14, 2011 #6
    I'm pretty sure you can factor a x^5 or greater polynomial out with a combination of rational zeros theorem, bounds on zeros theorem, factoring and long division. The book has been making me do that in this section.

    It would just be nice if it were cleaner.

    Well, I've only had to do one x^5 polynomial. But one can continue using those tools to get the answers.

    It would still be nice if this were a bit less messy though.
     
  8. Dec 14, 2011 #7

    I like Serena

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    Homework Helper

    How is it messy? :confused:

    Usually in exercises with difficult polynomials either 1, 0, or -1 used to be a root, so you could simplify.
    Now, you need to consider a few more possibilities.
    And if there are too many, you can shift and/or scale to reduce the number.
     
  9. Dec 14, 2011 #8

    Deveno

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    actually, it is a well-known theorem due to abel and ruffini, that for quintic polynomials and higher, NO general solution by the methods you suggest exists. this does not preclude SOME quintics from being solvable, but some are not.

    as an exercise, try to find a solution by radicals of x5-x-1.
     
  10. Dec 15, 2011 #9

    eumyang

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    What do you mean by solving? What you have is an expression, and you cannot solve an expression.

    Just because you have been solving quintics in the book does not mean that you can solve any quintic that's out there. Regarding lower degree polynomials, there exists a cubic formula and a quartic formula. You can find them easily enough through a search.
     
  11. Dec 15, 2011 #10

    You can solve it for zero, which is what is meant when typically when people speak of solving equations. You can also solve it for many other numbers.
     
  12. Dec 15, 2011 #11

    pwsnafu

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    What eumyang is saying is that in your post you didn't write down an equation. You wrote an expression. Equations must have an equal sign.

    You can however say
    "find the zeros of the polynomial 3x3+4x2-7x+2"
     
  13. Dec 15, 2011 #12

    Mark44

    Staff: Mentor

    No, as already mentioned, you can't solve an expression, and "solving it for zero" is meaningless. You can evaluate an expression at a particular value of the variable. Is this what you meant?

    And this is the same as saying "find the solutions of the equation 3x3+4x2-7x+2 = 0.
     
  14. Dec 15, 2011 #13

    Mark44

    Staff: Mentor

    Emphasis added.
    Any potential rational zeroes are listed by the theorem. It doesn't say anything at all about irrational zeroes.
     
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