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Finding zeros

  1. Jan 26, 2008 #1
    I haven't done this in ages, and I'm having trouble recalling how to factor a higher order polynomial. I almost always do this graphically, but for this case I'm interested in an algebraic solution. Specifically, I'm looking at ax + x^3 - x^5 = 0 (with a = an integer >0.)

    Clearly 0 is one solution since x(a + x^2 - x^4) =0. Is there a way to solve for the 2 others? Thanks.
  2. jcsd
  3. Jan 26, 2008 #2
    Generally i do not know how to find the roots of higher order polynomials myself, of an order n>=3. but in yor case it looks easy,because it is not of a complete form.
    like you said one trivial answer is x=0, the others will be the solutions to the equation
    a + x^2 - x^4=0, so u may want to take a substituton like this t=x^2, so you will end up with sth like this

    a+t-4t^2=0, now you now how to solve this right?
    and after that just go back and find solutions for x.
  4. Jan 26, 2008 #3


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    The remaining polynomial is quadratic in x2. Therefore you can get x2=(1+-(1+4a)1/2)/2. I'm sure you can finish.
  5. Jan 27, 2008 #4
    Thanks, I hadn't thought to substitute. I got it.
  6. Jan 28, 2008 #5


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    Note that it's a degree 5 polynomial, so you should have found 5 solutions.

    There are formulas for cubic and quartic equations, but I don't think anyone knows them by heart (ok, maybe a few, but certainly very few people). For [itex]n \ge 5[/itex] general solutions do not even exist (I believe this can be proven using ring theory).
  7. Jan 28, 2008 #6
    That's a bit of relief, i started feeling stupid for not being able to find general solutions for these higher order polynomials. For those of a degree of 3 i have seen some formulas, but havent actually tried to look closely what is going on. I am gonna have to have a look upon these things, cuz they come up sometimes in calculus one as well! And thnx, for enlightening me for polynomials of a degree of 5 or higher!
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