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Solving a 4-th degree polynomial

  1. Mar 22, 2014 #1
    1. The problem statement, all variables and given/known data

    This is part of a linear algebra problem. I've found the characteristic polynomial of a matrix
    however it's a degree 4 polynomial and I'm having trouble solving it

    2. Relevant equations

    λ43-3λ2-λ+2 = 0

    3. The attempt at a solution

    I replaced λ2 with a and did some manipulation, giving me 2 equations:

    a(a-3)+√a(a-1)=-2 and a(a-3)-√a(a-1)=-2

    (not even sure if this is correct though). By inspection I can see that 1 is one of the solutions for a is 1 hence making λ 1 and -1 but I need a more rigorous and formal way of solving this.
     
  2. jcsd
  3. Mar 22, 2014 #2

    ShayanJ

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    If you do the multiplication [itex] (ax^2+bx+c)(x+d)(x+e) [/itex], you will see the constant term is [itex] cde [/itex]. So if you decompose the constant term of the fourth degree polynomial in all possible factors(not talking about prime ones,just any kind of factor, even fractional ones), you can find some zeroes there, if the polynomial has real zeroes.
    I tried [itex] \lambda=-2 [/itex] and it is a solution. So you can divide the polynomial by [itex] \lambda+2 [/itex] and get a third degree polynomial. Now it is obvious that the other zero should be -1 and so you also can divide by [itex] \lambda+1 [/itex]. Now you have a second degree polynomial which is easy to solve. But I can readily say that the second degree factor is [itex] (\lambda-1)^2 [/itex]!
    Of course the order in which you use the zeros doesn't matter.
     
    Last edited: Mar 22, 2014
  4. Mar 22, 2014 #3

    mfb

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    It is possible to solve it in an analytical way, but usually this is not required for homework problems. Guessing zeros (based on factors of the absolute term) is perfectly acceptable.
     
  5. Mar 22, 2014 #4
    Thanks for the reply, I kinda know what you mean, but not really. I see how cde make the constant term. What do you mean by "real zeros"?
     
  6. Mar 22, 2014 #5

    ShayanJ

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    We have a kind of numbers called Real numbers. They're the numbers we usually work with. Examples are 1,2.64,6/5,[itex] -\pi [/itex].
    The equation [itex] x^2-\pi^2=0 [/itex] has zeroes [itex] \pm \pi [/itex]. But what about [itex] x^2+\pi^2=0 [/itex]?
    We have [itex] x^2=-\pi^2 [/itex] but you can't square a Real number and get a negative number,that's just impossible. But if we define [itex] i^2=-1 [/itex], then the equation [itex] x^2+\pi^2=0 [/itex] has imaginary zeroes, [itex] \pm i \pi [/itex].
    You can also have complex numbers which have the form a+bi for a and b being real.
     
  7. Mar 22, 2014 #6
    As the other guys said before me, in these excercises you have to guess a root of the polynomial. A nice theorem, if I remember correctly, is that if the polynomial has an integer root then it divides the last constant term. In this case, it's too. So, try by substituting λ=1,2,-1,-2. If that works, divide by λ-x, where x is the root of the polynomial.

    A nicer trick is "grouping". You group terms of the equation to form a sum of identical equations. For example, in this case we have:

    λ43-3λ2-λ+2=(λ43-2λ2)+(-λ2-λ+2)

    Can you go it any further? :)
     
  8. Mar 22, 2014 #7

    Borek

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    http://mathworld.wolfram.com/RationalZeroTheorem.html
     
  9. Mar 22, 2014 #8

    HallsofIvy

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    I'm not sure why you would replace [itex]\lambda^4[/itex] with [itex]a[/itex]. It should have been obvious from the start that [itex]\lambda= 1[/itex] is a root. And then, by regular division or "synthetic division" it should be clear that [itex]\lambda^4+ \lambda^3- 3\lambda^2- \lambda+ 2= (\lambda- 1)(\lambda^3+ 2\lambda^2- \lambda- 2)[/itex]. And, again, it is easy to check that [itex]\lambda= 1[/itex] is a root. So, dividing again, [itex]\lambda^4+ \lambda^3- 3\lambda^2- \lambda+ 2= (\lambda- 1)^2(\lambda^2+ 3\lambda+ 2)[/itex].

    And that last quadratic is easy to factor. The point of this exercise is: while there exist a "formula" for solving general fourth degree polynomial equations, it is extremely complicated and it is a good idea to start by using the "rational root theorem" or "rational zero theorem" that Borek gave a link to:

    "If a polynomial with integer coefficients has a rational number zero, a/b, then a must evenly divide the constant term and b must evenly divide the leading coefficient".

    Here the polynomial is [itex]\lambda^4+ \lambda^3- 3\lambda^2- \lambda+ 2[/itex] which has constant term 2 and leading coefficient 1 so any rational 0 must be an integer (denominator must divide 1 so can only be [itex]\pm 1[/itex]) that evenly divides 2. That is, the only possible rational roots are [itex]\pm 1[/itex] and [itex]\pm 2[/itex].

    There is no guarantee that a polynomial has any rational zeros but it is worth checking because it is so difficult to find non-rational zeros!
     
  10. Mar 22, 2014 #9

    SammyS

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    You have been given many helpful suggestions.

    If you have access to a graphing calculator, you could try graphing

    ##\displaystyle y = x^4+x^3-3 x^2-x+2\ .##

    The zeros of the polynomial are readily apparent. Further, check that the roots are truly what they appear to be.
     
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