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eunjii Group Title

Find the 3rd polynomial with roots 3 and 2 + i I know my roots are 3, 2 + i, 2 - i, but I don't know how to start my equation. Can someone please show me the steps to solve this problem?

  • one year ago
  • one year ago

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  1. satellite73 Group Title
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    you are looking for a quadratic whose zeros are \(2+i\) and \(2-i\) when you find that, you need to multiply it by \(x-3\) for your final answer there are a few ways to find the quadratic

    • one year ago
  2. satellite73 Group Title
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    one way is to memorize the following, if \(a+bi\) is a zero of a quadratic with leading coefficient 1, the quadratic is \[x^2-2ax+(a^2+b^2)\]

    • one year ago
  3. satellite73 Group Title
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    another way, in case you don't feel like memorizing that one, is to start with \[x=2+i\] and work backwards: \[x=2+i\] \[x-2=i\] \[(x-2)^2=i^2\] \[x^2-4x+4=-1\] \[x^2-4x+5=0\]so your quadratic is \(x^2-4x+5\)

    • one year ago
  4. eunjii Group Title
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    sorry I asked my question wrong. . .it was "Find the 3rd degree polynomial with roots 3 and 2 + i"

    • one year ago
  5. satellite73 Group Title
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    a final and not so pretty way is to start with \[(x-(2+i))(x-(2-i))\] and multiply out

    • one year ago
  6. satellite73 Group Title
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    yes, that is what we are trying to do

    • one year ago
  7. satellite73 Group Title
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    first step, find the quadratic with zeros \(2+i\) and its conjugate \(2-i\)

    • one year ago
  8. satellite73 Group Title
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    that will be \(x^2-4x+5\) no matter what method you use then your final answer will be \[(x-3)(x^2-4x+5)\]

    • one year ago
  9. eunjii Group Title
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    Thank you so much~<3 The explanation was very clear and helpful.

    • one year ago
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