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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?

Mathematics
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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 way is to memorize the following, if \(a+bi\) is a zero of a quadratic with leading coefficient 1, the quadratic is \[x^2-2ax+(a^2+b^2)\]
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 \(x^2-4x+5\)

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Other answers:

sorry I asked my question wrong. . .it was "Find the 3rd degree polynomial with roots 3 and 2 + i"
a final and not so pretty way is to start with \[(x-(2+i))(x-(2-i))\] and multiply out
yes, that is what we are trying to do
first step, find the quadratic with zeros \(2+i\) and its conjugate \(2-i\)
that will be \(x^2-4x+5\) no matter what method you use then your final answer will be \[(x-3)(x^2-4x+5)\]
Thank you so much~<3 The explanation was very clear and helpful.

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