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anonymous

  • one year ago

Using the given zero, find all other zeros of f(x). -2i is a zero of f(x) = x4 - 32x2 - 144

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  1. anonymous
    • one year ago
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    if \(2i\) is a zero then so is \(-2i\) therefore one factor is \(x+2i\) and another is \(x-2i\) multiply the factors, get \[x^2+4\]

  2. anonymous
    • one year ago
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    therefore this sucker factors as \[x^4-32x^2-144=(x^2+4)(\text{something})\]

  3. anonymous
    • one year ago
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    2i, 12, -12 2i, 6i, -6i 2i, 6, -6 2i, 12i, -12i

  4. anonymous
    • one year ago
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    on the other hand you can solve this directly without knowing any zeros, put \(u=x^2\) and solve \[u^2-32u-144=0\] by factoring

  5. anonymous
    • one year ago
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    those are my options

  6. anonymous
    • one year ago
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    once you find \(u\) then replace it by \(x^2\) and solve

  7. anonymous
    • one year ago
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    you know how to factor this ?

  8. anonymous
    • one year ago
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    hint, one of the factors is \(x^2+4\)

  9. anonymous
    • one year ago
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    is the answer a?

  10. anonymous
    • one year ago
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    probably want me to check?

  11. anonymous
    • one year ago
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    yes please :)

  12. anonymous
    • one year ago
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    actually i change my mind probably not

  13. anonymous
    • one year ago
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    \[x^4-31x-144=0\\ (x^2+4)(x^2+bx+c)=0\]can factor easily how many times does \(4\) go in to \(144\)?

  14. anonymous
    • one year ago
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    36

  15. anonymous
    • one year ago
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    actually \(-36\)

  16. anonymous
    • one year ago
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    so this factors as \[(x^2+4)(x^2-36)=0\]

  17. anonymous
    • one year ago
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    final job it to solve \[x^2-36=0\] for \(x\)

  18. anonymous
    • one year ago
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    6, -6

  19. anonymous
    • one year ago
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    yup

  20. anonymous
    • one year ago
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    Oh i see what i did wrong now, thanks!

  21. anonymous
    • one year ago
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    yw

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