Question

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?

Answer 1

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

Answer 2

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

Answer 3

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 \(x^2-4x+5\)

Answer 4

sorry I asked my question wrong. . .it was "Find the 3rd degree polynomial with roots 3 and 2 + i"

Answer 5

a final and not so pretty way is to start with
\[(x-(2+i))(x-(2-i))\] and multiply out

Answer 6

yes, that is what we are trying to do

Answer 7

first step, find the quadratic with zeros \(2+i\) and its conjugate \(2-i\)

Answer 8

that will be \(x^2-4x+5\) no matter what method you use
then your final answer will be
\[(x-3)(x^2-4x+5)\]

Answer 9

Thank you so much~<3 The explanation was very clear and helpful.