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anonymous

  • one year ago

Write a polynomial with rational coefficients having roots 3, 3+i, and 3-i. Part 1: Write the factors (in the form x-a) that are associated with the roots (a) given in the problem. Part 2: Multiply the 2 factors with complex terms to produce a quadratic expression. Part 3: Multiply the quadratic expression you just found by the 1 remaining factor to find the resulting cubic polynomial.

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  1. anonymous
    • one year ago
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    nah don' t do that

  2. anonymous
    • one year ago
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    \[(x-(3+i))(x-(3-i))\] is a pain to multiply there is a much easier wy

  3. anonymous
    • one year ago
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    you want a quadratic with zeros \(3+i\) and \(3-i\) there is an easy way to find it also a real real easy way which would you like to use?

  4. anonymous
    • one year ago
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    Well I am pretty sure I can find the easier way, but the packet I'm doing requires me to do it the hard way, unfortunately. Can you help me with the harder way?

  5. anonymous
    • one year ago
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    what difference does it make? you have to end up with a quadratic, might as well just write it

  6. anonymous
    • one year ago
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    Well each of the steps is graded separately as a different question. So if I skip the first two steps (or do them differently) then I get points taken off.

  7. anonymous
    • one year ago
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    if the zeros are \(a\pm bi\) the quardratic is \[x^2-2ax+(a^2+b^2)\]

  8. anonymous
    • one year ago
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    ok then lets multiply

  9. anonymous
    • one year ago
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    So how do I find my factors for this problem?

  10. anonymous
    • one year ago
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    if a root is \(r\) then a factor is \((x-r)\)

  11. anonymous
    • one year ago
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    your roots are 3, 3 + i, 3 - i so your factors are \[(x-3)(x-(3+i))(x-(3-i))\]

  12. anonymous
    • one year ago
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    Okay yes that's what I got

  13. anonymous
    • one year ago
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    that is the answer to A

  14. anonymous
    • one year ago
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    for B they want you to multiply \[(x-(3+i))(x-(3-i))\]

  15. anonymous
    • one year ago
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    So the complex roots only? I see.

  16. anonymous
    • one year ago
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    that is what is says right?

  17. anonymous
    • one year ago
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    Yep, thanks.

  18. anonymous
    • one year ago
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    Do I multiply this by FOIL?

  19. anonymous
    • one year ago
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    i wouldn't

  20. anonymous
    • one year ago
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    well ok i guess, but keep \(3+i\) and \(3-i\) together lets do the first outer inner last business, but to it the easy way

  21. anonymous
    • one year ago
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    so don't distribute the x to the roots?

  22. anonymous
    • one year ago
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    first is a no brainer, it is \(x^2\)

  23. anonymous
    • one year ago
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    lets skip to the last when you multiply a complex number \(a+bi\) by its complex conjugate \(a-bi\) you get the real number \(a^2+b^2\) to the "last" is \[3^2+1^2=10\]

  24. anonymous
    • one year ago
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    Oh okay I've never seen that before.

  25. anonymous
    • one year ago
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    then "outer and inner"\[-(3+i)x-(3-i)x\]combine like terms, the \(i\) part goes bye bye and you are left with \(-6x\)

  26. anonymous
    • one year ago
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    So did you distribute the x first, then subtract?

  27. anonymous
    • one year ago
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    if you haven't seen that then you should learn it now \[(a+bi)(a-bi)=a^2+abi-abi-b^2i^2=a^2+b^2\]

  28. anonymous
    • one year ago
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    yes i would call it "distribute then combine like terms" but it is the same thing

  29. anonymous
    • one year ago
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    Okay thanks. Okay and so the last part is to multiply the quadratic we found by x-3?

  30. anonymous
    • one year ago
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    final answer \[x^2-6x+10\] as promised

  31. anonymous
    • one year ago
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    yes i will let you do that yourself, it is easy enough right?

  32. anonymous
    • one year ago
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    Yeah it'll take me a while because I'm rusty but I can do it. Thank you so much!

  33. anonymous
    • one year ago
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    yw

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