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Help with finding Zeros of a polynomial with 1+i as a zero

  1. May 28, 2016 #1
    1. The problem statement, all variables and given/known data
    p(x) = x^3 − x^2 + ax + b is a real polynomial with 1 + i as a zero, find a and b and find all of the real zeros of p(x).


    3. The attempt at a solution

    1-i is also a zero as it is the conjugate of 1+i

    so

    (x-(1+i))(x-(1-i))=x^2-2x+2

    let X^3-x^2+ax+b=x^2-2x+2(ax+d)

    constant terms: b=2d, d=b/2

    x terms: -2d+2a, a=2d

    This is what I have done so far but I am unsure how to find a and b and the third zero

    Any Help would be appreciated
     
  2. jcsd
  3. May 28, 2016 #2

    PeroK

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    You've lost me here. That last step makes no sense.

    ##p(x) = (x^2 -2x + 2)(x + d)##

    Surely?
     
  4. May 28, 2016 #3
    That is the third root which I have to find

    How else would I go about solving it?
     
  5. May 28, 2016 #4

    PeroK

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    What's the most obvious thing you can do with

    ##(x^2 -2x + 2)(x + d)##
     
  6. May 28, 2016 #5
    I can Expand it which would give me

    x^3+dx^-2x^2-2xde+2x+2d

    What would I do now?

    Would I sub in a (1+i) in for x which would give me d2i-2d+2id+2d
     
  7. May 28, 2016 #6
    It should say x^3+dx^2-2x^2-2xd+2x+2d
     
  8. May 28, 2016 #7
    Hi
    I have a very easy method, and i am gonna give you a hint.

    Assume the third root to be C

    And then use the relation between the coeffecients of the equation and the sum taken one at a time,two at a time , three at a time.
     
  9. May 28, 2016 #8

    pasmith

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    It is more natural to write [itex]x^3 - x^2 + ax + b = (x^2 - 2x + 2)(x - x_0)[/itex] since the root we seek is then [itex]x_0[/itex] rather than [itex]-d[/itex].

    You have three unknowns. You need therefore three simultaneous equations which they must satisfy. You can obtain these by setting [tex]
    x^3 - x^2 + ax + b = (x^2 - 2x + 2)(x + d)[/tex] and comparing coefficients of powers of [itex]x[/itex].
     
  10. May 28, 2016 #9
    using constant term
    b=2d

    x terms

    a=2d+2

    x^2 term

    -1=d-2

    Would this be the equations that I need to use?
     
  11. May 28, 2016 #10
    did you see post #7 ? use that its much quicker.
     
  12. May 28, 2016 #11
    I'm not sure that I understand would I just use C as a root and expand (1+i)(1-i)C
     
  13. May 28, 2016 #12
    Nope.

    1st root :- 1+i
    2nd root:- 1-i
    3rd root :- C

    Sum of roots = -(coeff of x^2)/(coeff of x^3)
    Sum of product of the roots taken 2 at a time= (coeff of x)/(coeff of x^3)
    Sum of product of the roots taken 3 at a time = -(constant)/(coeff of x^3)

    EDIT:

    Capture.jpg
     
  14. May 28, 2016 #13
    Sum of roots taken 1 at a time = -(coeff of x^2)/(coeff of x^3) =-1/1=1
    Sum of roots taken 2 at a time= (coeff of x)/(coeff of x^3) = a/1=a
    Sum of roots taken 3 at a time = -(constant)/(coeff of x^3) = b/1=b

    What would I do now?
     
  15. May 28, 2016 #14
    I edited it, refresh and check.

    your third equation should be= '-b' not 'b'

    and if r1,r2,r3 are the roots of 3rd degree polynomial
    then
    sum of product of roots taken 2 at a time means
    r1.r2+ r2.r3 + r3.r1

    rest you can understand.
     
  16. May 28, 2016 #15

    PeroK

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    So, you have:

    ##p(x) = x^3 - x^2 +ax + b = x^3 + (d-2)x^2 + (2-2d)x + 2d##

    It shouldn't be hard to solve for ##d, a, b## from there!
     
  17. May 28, 2016 #16
    does this meant that

    -x^2=(d-2)x^2
    -1=d-2
    d=1
     
  18. May 28, 2016 #17

    PeroK

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    Yes! Just compare coefficients, as was mentioned in post #8.
     
  19. May 29, 2016 #18

    epenguin

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    The way to find a simpler (reduced degree) equation from a polynomial equation of which you already know some roots (i.e. you know a factor of the polynomial) is to divide the overall polynomial by the polynomial that you know.

    Equivalently and I think this way is easier here, eliminate between the two polynomials until you get a linear factor in this case.

    In this way you should get an equation linear in x. You then identify coefficients.

    If you are handy with determinants there is another, still equivalent, way to do it using them. Which I tried – but it seems I am not as handy as I thought. :redface:
     
    Last edited: May 29, 2016
  20. May 29, 2016 #19

    SammyS

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    It seems to me that polynomial division is the most direct way to get the complete solution. The real zero "pops out" immediately. Along with this, one simply chooses a and b to make the remainder be zero.

    Unfortunately, this thread has been marked as "Solved" despite the fact that OP has not given his/her solution. Apparently OP has never come across the epenguin's signature message !
     
  21. May 29, 2016 #20
    Since the thread has been marked solved :-

    1+i+1-i+C=1 => C=-1

    (1+i)(1-i) + (1-i)(C) + (1+i)(C) = a => 2 +2C=a => a=0.

    (1+i)(1-i)(C) = -b => -b=2C => b=2.

    All the roots : 1+i , 1-i , -1
    a: 0
    b: 2

    The equation becomes:- ## x^3-x^2+2 =0##

    ## (x+1)(x^2-2x+2)=0##
     
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