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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. 4, -2, and -1 + 2i here are my choices f(x) = x^4 - 7x^2 - 26x - 40 f(x) = x^4 - 3x^3 - 8x^2 - 13x - 40 f(x) = x^4 + 6.5x^2 - 26x - 40 f(x) = x^4 - 3x^3 + 8x^2 + 13x + 40

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  1. anonymous
    • one year ago
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    this again ready?

  2. anonymous
    • one year ago
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    i dont know how to do this at all lolol

  3. anonymous
    • one year ago
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    ok then we better go slow

  4. anonymous
    • one year ago
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    do you know if \(r\) is a zero, then one factor must be \((x-r)\)? so in this case you have \(4\) is a zero making one factor \((x-4)\)

  5. anonymous
    • one year ago
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    clear or not? that is really what we need to start this

  6. anonymous
    • one year ago
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    oh okay so (x-4) and (x+2) ?

  7. anonymous
    • one year ago
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    right

  8. anonymous
    • one year ago
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    we start with \[(x-4)(x+2)\] now we have more work to do

  9. anonymous
    • one year ago
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    you also need the quadratic that has the zeros of \(-1+2i\) and it "conjugate" \(-1-2i\) next job is to find that

  10. anonymous
    • one year ago
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    any ideas? "no" is a fine answer

  11. anonymous
    • one year ago
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    no i dont know

  12. anonymous
    • one year ago
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    ok there is a hard way, an easy way, and a really really easy way lets do the easy way first, then the real easy way

  13. anonymous
    • one year ago
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    work backwards starting with \[x=-1+2i\] add \(1\) to get \[x+1=2i\]

  14. anonymous
    • one year ago
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    then square both sides (carefully) to get \[(x+1)^2=(2i)^2\] or \[x^2+2x+1=-4\]

  15. anonymous
    • one year ago
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    add 4 to both sides to finish with \[x^2+2x+5\]

  16. anonymous
    • one year ago
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    final job is to multiply \[(x-4)(x+2)(x^2+2x+5)\]i would cheat so as not to screw up the algebra

  17. anonymous
    • one year ago
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    http://www.wolframalpha.com/input/?i=%28x-4%29%28x%2B2%29%28x^2%2B2x%2B5%29

  18. anonymous
    • one year ago
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    and would that be the answer? or is there more

  19. anonymous
    • one year ago
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    OOO got it. thank you

  20. anonymous
    • one year ago
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    yeah that is it

  21. anonymous
    • one year ago
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    yw

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