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anonymous

  • one year ago

Write a polynomial function of minimum degree with real coefficients whose zeros include those listed. Write the polynomial in standard form. 7, -11, and 2 + 6i

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  1. anonymous
    • one year ago
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    in order to have real coefficients, all complex roots must appear as conjugates

  2. anonymous
    • one year ago
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    do you know what the conjugate of 2+6i is?

  3. anonymous
    • one year ago
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    no :(

  4. anonymous
    • one year ago
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    -2-6i

  5. anonymous
    • one year ago
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    http://www.mathsisfun.com/numbers/complex-numbers.html a+bi and a-bi are complex conjugates

  6. anonymous
    • one year ago
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    ohh ok

  7. anonymous
    • one year ago
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    so 2-6i

  8. anonymous
    • one year ago
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    yep so can you write the polynomial in factored form?

  9. anonymous
    • one year ago
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    7, -11, and 2 + 6i

  10. anonymous
    • one year ago
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    no... in factored form it would be something like this... f(x) = (x-c)(x-d)(x-(a+bi))(x-(a-bi))

  11. anonymous
    • one year ago
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    f(x)=(x+7)(x -11) (2 + 6i)

  12. anonymous
    • one year ago
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    not quite... the complex factor would be x-(2+6i) and you would need its conjugate as a factor also, if your polynomial is to have real coefficients

  13. anonymous
    • one year ago
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    i don't really understand

  14. anonymous
    • one year ago
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    if 4 is a root, it implies that f(4) = 0 and that the polynomial must have a factor of x-4, correct?

  15. anonymous
    • one year ago
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    yes

  16. anonymous
    • one year ago
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    so if 2+6i is a root, then x-(2+6i) is a factor. but if your polynomial is to have real coefficients then all complex roots must occur as complex conjugate pairs

  17. anonymous
    • one year ago
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    so x-(2-6i) also must be a factor

  18. anonymous
    • one year ago
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    once you have the factored form, just multiply out to get into standard form

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