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anonymous

  • 5 years ago

Find a polynomial function of least degree having only real coefficients with zeros as given. -3,2,-i, and 2+i

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  1. anonymous
    • 5 years ago
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    mrsheinold, if these are the zeros of your polynomial P(z) for z complex, then you have\[P(z)=(x+3)(z-(2-i))(z-(2+i))\]All you need to do is expand this out to get your polynomial in a more amenable form:\[P(z)=z^3-z^2-7z+15\]

  2. anonymous
    • 5 years ago
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    You'll always have real coefficients if your zeros are real or come in complex conjugate pairs, like you have...

  3. anonymous
    • 5 years ago
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    Ok... I dont understand. The example in the book uses f(x) not P(z) im assuming thats ok.. but when I got my formulas at first I got f(x)=(x+3)(x-2)(x+i)(x-2-i) how did you get your first formula??

  4. anonymous
    • 5 years ago
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    f(x), P(z)...semantics, don't worry about it. Use what the book uses. You get the first formula by the following: if \[\alpha, \beta, \gamma\]are roots of a polynomial, f(x), then \[f(x)=(x-{\alpha})(x-\beta)(x-\gamma)\]All you need to do is sub. your roots in.

  5. anonymous
    • 5 years ago
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    They are defined as roots of the equation because when x takes any of those values, one of the factors becomes zero, giving f(x)=0.

  6. anonymous
    • 5 years ago
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    Is this clearer?

  7. anonymous
    • 5 years ago
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    Not really. I have 4 numbers -3,2,-i, 2+i I'm suppose to change the signs and put them in () So I thought -3 would change to (x+3) and the 2 would change to (x-2) -i would change to (x+i) and 2+i would change to (x-2-i) are those right or wrong??

  8. anonymous
    • 5 years ago
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    Oh sorry bud, I read it as 2-i, not 2,-i...like I said, I'm ready for bed. Hang on...

  9. anonymous
    • 5 years ago
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    mrsheinold, since you are given 4 numbers two of which are complex, your desired polynomial should have 6 factors, P(x)=(x+3)(x-2)(x-i)(x+i)(x-(2+i))(x-(2-i)) as stated above when you are given roots of a polynomial that are complex, those roots will appear in conjugate pairs. e.g. find the roots of P(x)=x^2+1

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