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BBB83

  • 2 years ago

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

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

  3. dumbcow
    • 2 years ago
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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

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

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

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

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

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

  9. BBB83
    • 2 years ago
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    177x?

  10. dumbcow
    • 2 years ago
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    yep:)

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

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