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Homework Help: Polynomial Help

  1. Oct 2, 2005 #1
    If m, n, and 1 are non-zero roots of the equation [tex]x^3 - mx^2 + nx - 1 = 0[/tex], then find the sum of the roots

    This is what I did..

    m, n, 1 are the roots. m and n not equal to 0

    [tex]x^3 - mx^2 + nx - 1 = 0[/tex]
    f(m) = 0 --> m^3 - m^3 + mn - 1
    1 = mn (1)

    f(1) = 1 - m + n - 1 = 0
    ... m = n (2)

    Sub (2) --> (1)

    [tex]m^2 = 1 [/tex]
    [tex]m = +/- 1

    since m and n are equal... the roots must be either 1, 1, 1 or -1, -1, 1. The answer on the sheet says the answer is -1. So my question is, how do we determine which roots are the answers. Thanks
     
  2. jcsd
  3. Oct 2, 2005 #2

    hotvette

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    This might work. If m, n, and 1 are the roots, that means (x-m)*(x-n)*(x-1) = 0. If you multiply this out and substitute the 2 different sets of values for m & n, I'm guessing that only 1 set is valid.
     
    Last edited: Oct 2, 2005
  4. Oct 2, 2005 #3
    oops nvm it didn't work
     
    Last edited: Oct 2, 2005
  5. Oct 8, 2005 #4

    hotvette

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    looks like there are 2 possible solutions.
     
  6. Oct 8, 2005 #5

    AKG

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    There is only one solution. I used long division. If you divide your polynomial by (x-1), you know you should get a quadratic polynomial, but you get:

    x² + (1-m)x + (n+1-m) + (n-m)/(x-1)

    So you know n - m = 0. This is one useful fact. Now you're left with:

    x² + (1-m)x + (n+1-m)

    You know that m is a root of this polynomial, so dividing by (x-m) should give a linear polynomial. When you do the long division, you get:

    x + 1 + (n+1)/(x-m)

    So you know n + 1 = 0. This gives n = -1. Having already got that n - m = 0, you now get m = -1. And you have the other root being 1.

    (-1) + (-1) + 1 = -1.
     
  7. Oct 8, 2005 #6

    hotvette

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    Good one AKG! :redface:
     
  8. Oct 9, 2005 #7

    HallsofIvy

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    In your very first post, you note that the solutions must be either {1, 1, 1} or {1, -1, -1} and ask how can you tell that only {1, -1, -1} is correct.

    If {1, 1, 1} were the roots, then the equation could be written
    (x-1)(x-1)(x-1)= 0 but (x-1)(x-1)(x-1)= x3- 3x2+ 3x- 1, not x3- x2+ x- 1 as it should be with m=n= 1.

    If {1, -1, -1}, on the other hand, were the roots then the equation could be written (x-1)(x+1)(x+1)= (x2- 1)(x+1)= x3+x2- x- 1 as required.
     
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