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BBB83

form a third degree polynomial function with real coefficients such that 8+i and 7 are zeros f(x) = ???

  • 2 years ago
  • 2 years ago

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  1. rivermaker
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    Since coefficients are real, the two complex roots have to be conjugates. So the simplest polynomial is \[(x-7)(x -8 -i)(x-8+i)\]

    • 2 years ago
  2. BBB83
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    The question wants the answer in a X^3 +bx^2+cX+# ?? so im confused.....

    • 2 years ago
  3. dumbcow
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    x = 8 +- i x-8 = +-i (x-8)^2 = -1 (x-8)^2 +1 = 0 x^2-16x +65 = 0 --> (x-7)(x^2-16x+65) Now distribute

    • 2 years ago
  4. jerwyn_gayo
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    @dumbcow, why -i if the given is positive(+)?

    • 2 years ago
  5. dumbcow
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    because complex zeros always come in pairs

    • 2 years ago
  6. jerwyn_gayo
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    ahh, yeah. . (+) and (-)

    • 2 years ago
  7. BBB83
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    x^3-23x^2-57x-455? is that right?

    • 2 years ago
  8. dumbcow
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    everything except the "-57x" term

    • 2 years ago
  9. BBB83
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    177x?

    • 2 years ago
  10. dumbcow
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    yep:)

    • 2 years ago
  11. BBB83
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    thanks for the direction much appreciated

    • 2 years ago
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