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Solving the polynomial

  1. Sep 2, 2007 #1
    1. Solve the eqution: x^4-2x^3-6x^2-2x+1=0

    2. Relevant equations
    Hint: let v=x+1/x

    3. The attempt at a solution
    Well, I didvided by x^2:
    since v=x+1/x, -2x-2/x=-2v

    So now I have: 2v=x^2-2x-6 but I really don't know what to do with this, I tried rearranging to find x in terms of v to substitute back into the original equation but that got quite complicated and messy!

    The back of the book says the solutions are: -1,2+3^0.5, 2-3^0.5
    Last edited: Sep 2, 2007
  2. jcsd
  3. Sep 2, 2007 #2
    Don't divide with [itex]x^2[/tex]. Try to guess a solution (test 0, 1, -1 etc.) and then perform polynomial division (if you know how).
  4. Sep 2, 2007 #3
    erm ok thanks-so am I trying to divide it so I get x+1/x ? How do I figure out what to divide by?
    Last edited: Sep 2, 2007
  5. Sep 2, 2007 #4
    Divide until you get a quadratic. Try inputing easy numbers like -1,0,1 into the equation and if it equals 0 then it is a factor. In this quartic (x+1) ends up being a factor twice (I got a different answer from the book.) So that means that -1 is a factor. Then use Synthetic Division (Which in my opinion is much easier and will produce less mistakes than polynomial division) and keep on trying until you get down to a quadratic and just do the quadratic formula. (-b +- sqrt(b^2 - 4ac) / 2a)
  6. Sep 2, 2007 #5
    What shall I divide by to get a quadratic? I've tried dividing by x, x^2 and x^3
  7. Sep 2, 2007 #6
    i m tryin this way:

    seems like an elegant eq~!
    Last edited: Sep 2, 2007
  8. Sep 2, 2007 #7
    I'm now trying to factorise the quartic into two quadratics like rootX says to do but how are we supposed to know what goes into the 1st bracket? Do we have to find the factors using the synthetic division (do we just guess at what coud be a factor and see if there is no remainder?)
  9. Sep 2, 2007 #8


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    if [tex]v=x+\frac{1}{x}[/tex] consider what [tex]v^2[/tex] will be and from the equation you have to solve...factorise out [tex]x^2[/tex] from it and see what happens
  10. Sep 2, 2007 #9


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    That's not right. Didn't you mean
    x^2 - 6 + 1/x^2 = 2v?

    And since the whole point of a change of variable is to try and express everything in terms of v you should consider higher powers of v to see if you can do anything with them
  11. Sep 2, 2007 #10
    ooh! I think I get it! Thanks for your help guys! I tried x=1 which gave a answer of 0 so that meant x+1 was a factor so then I used the polynomial division suggested by Moridin to cancel the quartic down then did the same again to get a quadratic and used the formula (thanks to SnipedYou) which gave the solutions.
    Is there a way to find a factor to start off without "guessing"?
  12. Sep 2, 2007 #11


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    There's a problem... 1 is not a solution of
    x^4 - 2x^3 - 6x^2 - 2x + 1=0​

    Surely you've seen the rational root theorem, which tells you how to write down a short list of rational numbers, in which every rational root (if any) appears?
  13. Sep 2, 2007 #12


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    By the way, do you know why you got the hint to try
    v = x + 1/x?​
    And do you know what would lead someone to think about trying that?
  14. Sep 3, 2007 #13


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    Hint for Hurkyl's question: symmetry.
  15. Sep 3, 2007 #14
    Uses Des Cartes Rule of Signs and then use the Rational Root Theorm
  16. Sep 3, 2007 #15


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    Divide both sides, as you say you did, by x^2:

    x^2- 2x- 6- 2/x+ 1/x^2= 0

    If v= x+ 1/x then v^2= x^2+ 2+ 1/x^2.

    Isn't it obvious to write the equation as

    (x^2+ 2+ 1/x^2)- 2(x+1/x)- 8= 0
    v^2- 2v- 8= 0. That factors easily.
  17. Sep 3, 2007 #16
    Oops sorry! I meant I'd used x=-1 not x=1. I've never heard of the rational root theorem but I'll look it up now. I get how to use the hint they've given now- thanks!
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