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Math Help - factorising another polynomial over C

  1. #1
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    Gold Coast, Queensland, Australia
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    factorising another polynomial over C

    Hi, I'm stuck on this one. I don't know why though. The answer in the book is 8.
    if P(z) = z^3 + 2z^2 - 6z + a and  P(1 - i) = 0, then a is equal to?
    A) 4 B) -6 C) 8 D) 6 E) 5

    So the first thing I saw was that there is a conjugate zero also so it looks like this
     P(z) = (z - 1 - i) (z - 1 + i) (z - z_{3})
    So then I thought i would expand out just the z's, not the z^2s etc. and let them equal the z term in the original expression, -6z
    z(-1 + i) + z(-z_{3}) + z(-1 - i) + z(-z_{3}) + z(-1 - i) + z(-1 + i) = -6z
    the z cancels out
    -1 + i - z_{3}  -1 - i -z_{3} - 1 - i - 1 + i = -6
    -4 - 2z_{3} = -6
    -2z_{3} = -2
    z_{3} = 1

    So then I now have (z - 1 - i)(z - 1 + i)(z - 1) and I will just expand our the constants to get the a
    But this doesn't seem to work. I won't bother writing it down because I have done it about 5 times and I can't seem to get the answer. Does anyone know a method to get this one out?
    David.
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  2. #2
    Member
    Joined
    Mar 2009
    From
    Gold Coast, Queensland, Australia
    Posts
    105
    I worked it out now.
    Thanks.
    I put it in the form  P(z) = (z - 1 + i)(z^2 + pz + q) and worked out p based on the z^2 term and then worked out q based on the p and the z term, and then allowed me to work out what a is.
    Thanks.
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