Question

make a polynomial where
1. the leading coefficient is one
2. the others are real and rational
3. the zeros are 3i and 2-i

Answer 1

ok ready?

Answer 2

yes

Answer 3

i agree

Answer 4

one zero is \(3i\) that makes another zero its conjugate \(-3i\) so the quadratic will be \[(x-3i)(x+3i)=x^2+9\]

Answer 5

another way we could have found \(x^2+9\) is to set \[x=3i\] square both sides and get \[x^2=-9\] then add 9 to get \[x^2+9\]
that takes care of the \(3i\) part

Answer 6

i forgot about congegents lol

Answer 7

we still have to deal with the \(2+i\) part

Answer 8

there is a hard way to find the quadratic with zeros at \(2\pm i\), an easy way, and a real real easy way
you pick

Answer 9

r r easy

Answer 10

lets do it the easy way first
the real real easy way requires memorizing something (as with most math, the more you know the easier things are)

Answer 11

set \[x=2+i\] and work backwards
subtract 2 get \[x-2=i\] then square both sides (carefully) get \[(x-2)^2=i^2\] or \[x^2-4x+4=-1\]

Answer 12

add 1 and get \[x^2-4x+5=0\] so your quadratic is \[x^2-4x+5\]

Answer 13

the real real easy way is to know that if the solution is \(a+bi\) then the quadratic is \[x^2-2ax+a^2+b^2\] in your case \(a=2,b=1\) quadratic is \[x^2-2\times 2x+2^2+1^2=x^2-4x+5\]

Answer 14

you still have one more job, to multiply \[(x^2+9)(x^2-4x+5)\] i will leave that to you

Answer 15

hope it was more or less clear, if you have a question or two let me know

Answer 16

thanks

Answer 17

yw