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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. 5, -3, and -1 + 2i

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  1. anonymous
    • one year ago
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    sorry it went away

  2. anonymous
    • one year ago
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    oh ok

  3. anonymous
    • one year ago
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    before we begin is it clear that two factors are \((x-5)\) and \((x+3)\)?

  4. anonymous
    • one year ago
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    yes

  5. anonymous
    • one year ago
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    ok so the real job (before multiplying out) is the find the quadratic polynoimial with zeros at \(-1+2i\) and its conjugate \(-1-2i\)

  6. anonymous
    • one year ago
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    you want the easy way, or the real real easy way?

  7. anonymous
    • one year ago
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    real real easy

  8. anonymous
    • one year ago
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    ok actually lets to the easy way first then the real real easy way

  9. anonymous
    • one year ago
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    we can work backwards starting with \[x=-1+2i\] add 1 and get \[x+1=2i\] then square (carefully) to get \[(x+1)^2=(-2i)^2\] or \[x^2+2x+1=-4\]

  10. anonymous
    • one year ago
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    add 4 to both sides and get \[x^2+2x+5=0\] and that is your polynomial

  11. anonymous
    • one year ago
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    the real real easy way requires memorizing something that if \(a+bi\) is a zero of a quadratic, then it is \[x^2-2ax+(a^2+b^2)\] so in your case \(a=-1,b=2\) and the quadratic is \[x^2-2\times (-1)x+(-1)^2+2^2\] i.e. \[x^2+2x+5\]

  12. anonymous
    • one year ago
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    those arent any options tho :(

  13. anonymous
    • one year ago
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    your final job is to multiply \[(x-5)(x+3)(x^2+2x+5)\]

  14. anonymous
    • one year ago
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    if it was me, i would cheat so as not to screw up the algebra when multiplying

  15. anonymous
    • one year ago
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    you know how to do that?

  16. anonymous
    • one year ago
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    wait let me try again

  17. anonymous
    • one year ago
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    ok i will leave you to it, then show you how to get the answer for sure

  18. anonymous
    • one year ago
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    yeah the answer i get does not match up with my choices

  19. anonymous
    • one year ago
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    http://www.wolframalpha.com/input/?i=%28x-5%29%28x%2B3%29%28x^2%2B2x%2B5%29

  20. anonymous
    • one year ago
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    see if that one does

  21. anonymous
    • one year ago
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    Oh thank-you! maybe i was multiplying something wrong!

  22. anonymous
    • one year ago
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    that is why i said "cheat" is is easy to make a mistake when multiplying all this muck out

  23. anonymous
    • one year ago
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    yw

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