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Math Help - Factorising a polynomail over c

  1. #1
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    Factorising a polynomail over c

    If z - 3i is a factor of P(z) = 2z^4 - 4z^3 + 21z^2 - 36z + 27, find the remaining factors.

    So I did this:
    Step 1: if z - 3i is a factor, then z + 3i is a factor.
    Step 2: P(z) = (z - z_{1})(z - z_{2})(z - z_{3})(z - z_{4})
    Step 3: P(z) = (z - z_{1})(z - z_{2})(2z^2 + pz +q)
    Step 4: 2z^4 - 4z^3 + 21z^2 - 36z + 27 = (z - 3i)(z + 3i)(2z^2 + pz +q)
    Step 5: 2z^4 - 4z^3 + 21z^2 - 36z + 27 = (z^2 + 9)(2z^2 + pz + q)
    Step 6: 9q = 27
    Step 7: q = 3
    Step 8: 9p = -36
    Step 9: p = -4
    Step 10: Find the factors from the quadratic. 2z^2 - 4z + 3
    Step 11: 2(z^2 - 2z + \frac{3}{2})
    Step 12: 2[(z^2 - 2z + 1) + \frac{3}{2} - \frac{2}{2})]
    Step 13: 2[(z - 1)^2 + \frac{1}{2})]
    Step 14: 2[(z - 1)^2 - (\frac{i}{\sqrt{2}})^2)]
    Step 15: 2[(z - 1 + \frac{i}{\sqrt{2}})(z - 1 - \frac{i}{\sqrt{2}})]
    So this is the answer I got, but the book says just that these are the two remaining factors without the leading 2, as follows:
    (z - 1 + \frac{i}{\sqrt{2}})(z - 1 - \frac{i}{\sqrt{2}}

    So did I get it wrong or is the 2 not really classified as part of the factor for some reason? I can't understand why the 2 isn't included in the answer in the back of the book.

    David.
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  2. #2
    No one in Particular VonNemo19's Avatar
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    I think it's because 2 doesn't qualify as a root.
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  3. #3
    Super Member Quacky's Avatar
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    Your working is absolutely fine. The 2 is omitted as it's a constant term. I've struggled to explain why for a few minutes, with absolutely no success, however. I'll think about it, or perhaps else will help whilst I try to find the correct wording.

    Edit: No need, courtesy of VonNemo
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  4. #4
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    \dfrac{z^4-4z^3+21z^2-36z+27}{z^2+9}=2z^2-4z+3.
    Now find two roots of 2z^2-4z+3.
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  5. #5
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    Thanks.
    I will take it that I factorised it correctly then.
    David.
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