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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?
 one year ago
 one year ago
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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satellite73Best ResponseYou've already chosen the best response.1
you are looking for a quadratic whose zeros are \(2+i\) and \(2i\) when you find that, you need to multiply it by \(x3\) for your final answer there are a few ways to find the quadratic
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
one way is to memorize the following, if \(a+bi\) is a zero of a quadratic with leading coefficient 1, the quadratic is \[x^22ax+(a^2+b^2)\]
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
another way, in case you don't feel like memorizing that one, is to start with \[x=2+i\] and work backwards: \[x=2+i\] \[x2=i\] \[(x2)^2=i^2\] \[x^24x+4=1\] \[x^24x+5=0\]so your quadratic is \(x^24x+5\)
 one year ago

eunjiiBest ResponseYou've already chosen the best response.0
sorry I asked my question wrong. . .it was "Find the 3rd degree polynomial with roots 3 and 2 + i"
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
a final and not so pretty way is to start with \[(x(2+i))(x(2i))\] and multiply out
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
yes, that is what we are trying to do
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
first step, find the quadratic with zeros \(2+i\) and its conjugate \(2i\)
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
that will be \(x^24x+5\) no matter what method you use then your final answer will be \[(x3)(x^24x+5)\]
 one year ago

eunjiiBest ResponseYou've already chosen the best response.0
Thank you so much~<3 The explanation was very clear and helpful.
 one year ago
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