Question

given a polynomial has roots 2i + 3 and 2i -3 find p(x)

Answer 1

is it really \(2i+3\) and \(2i-3\) ? very tricky if it is

Answer 2

yes

Answer 3

ok lets write both of these numbers in standard form first
they are \(3+2i\) and \(-3+2i\)

Answer 4

if \(3+2i\) is a zero of a polynomial with real coefficients, then so is its conjugate \(3-2i\)
now we can find a quadratic with those zeros

Answer 5

the hard way is to write in factored form as \((x-(3+2i))(x-3(3-2i))\) and multiply out
the easier way is to work backwards and put
\[x=3+2i\\
x-3=2i\\
(x-3)^2=-4\\
x^2-6x+9=-4\\
x^2-6x+13=0\] and that is your quadratic with the two zeros \(3\pm2i\)

Answer 6

the real easy way is to memorize that if \(a+bi\) is a zero of a quadratic, then the quadratic is
\[x^2-2ax+(a^2+b^2)\] but that requires memorizing it
in you case you would get
\[x^2-2\times 3x+(3^2+2^2)=x^2-6x+13\] as before

Answer 7

now we have to repeat the process with \(-3+2i\)
this time we get
\[x^2+6x+14\]

Answer 8

and finally your last job is to multiply
\[(x^2-6x+14)(x^2+6x+14)\]

Answer 9

the answer i got was way out of pocket