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Find roots to quintic polynomial

  1. Mar 23, 2009 #1
    1. The problem statement, all variables and given/known data
    Find the roots to the polynomial x4+x+1

    this is a problem I am stuck with for my algebra class

    2. Relevant equations
    there are a few tricks I have learned for finding roots like taking conjugates and rewriting using fundamental theorem of algebra but we haven't worked on this stuff too much


    3. The attempt at a solution

    by fundamental theorem of algebra, we can write x4+x+1 = (x-a)(x-b)(x-c)(x-d) for some complex numbers a,b,c,d

    so the roots a,b,c,d have to follow the equations:
    a + b + c + d = 0
    ad + ab + bd + cd + bc = 0
    abd + bdc + acd + abc = 1
    abcd = 1

    other than that I do not know what else to do to find the roots
     
  2. jcsd
  3. Mar 23, 2009 #2

    Mark44

    Staff: Mentor

    Any complex numbers that are solutions show up in pairs as conjugates a + bi and a - bi.

    Also, this is a quartic (degree 4), not a quintic (degree 5).
     
  4. Mar 23, 2009 #3
    Yup the part about conjugates is one of the only things I know about this subject
    and thanks, I put down quintic when I should have said quartic.
     
  5. Mar 23, 2009 #4

    lanedance

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    Homework Helper

    have a look at your equation, can you convince yourself there are no real roots? (ie is f(x) > 0 for all x in the reals?

    if so as Mark implied, you know there will be 2 sets of complex conjugate roots, try inputting these in your equation and solving
     
  6. Mar 23, 2009 #5
    okay so the roots are of the form: r1 = a+ib, r2 = a-ib, r3 = c+id, r4 = c-id where a,b,c,d are reals.

    then we obtain the equations 2a + 2c = 0 and
    (a2+b2)(c2+d2) = 1
     
  7. Mar 23, 2009 #6

    lanedance

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    on the right track, but you should get an equation for each coefficient of x
     
  8. Mar 23, 2009 #7
    okay so the equations are:
    2a + 2c = 0
    (a2 + b2) (c2 + d2) = 1
    2(a2c + ac2 + ad2+b2c) + i (a2b + d2b + c2b+b2d + a2d + c2d + d3 + b3) = 0
    a2 + b2 + c2 + d2 + 3ac + bc -i (ad + ac) = 1
     
  9. Mar 23, 2009 #8

    lanedance

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    multiple the conjugate pairs first

    for a complex number z = a + ib
    z.z* = (a^2 + b^2) is real
    z+z* = 2a is real
    so (x-z)(x-z*) = x^2 - (z+z*)x + z.z* is then real

    so there should be no coefficents of i in your equations, and there should be no d3 as it only appears in factors twice...
     
  10. Mar 23, 2009 #9
    I always seem to get coefficients of i in the equations though
     
  11. Mar 23, 2009 #10

    lanedance

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    Homework Helper

    try this
    for
    z = a + ib, z* = a-ib
    y = c + id, y* = c-id

    (x-z)(x-z*)(x-y)(x-y*) = (x^2 - (z+z*)x + z.z*)(x^2 - (y+y*)x + y.y*)

    there is clearly no i's left on the right hand side
     
    Last edited: Mar 24, 2009
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