Question

n=4; 2i and 3i are zeros; f(-2)=104

Answer 1

you want a fourth degree polynomial with these zeros?

Answer 2

start with
\[a(x+2i)(x-2i)(x+3i)(x-3i)\] and multiply out

Answer 3

yes how do i that?

Answer 4

not as bad as it seems since
\[(x+2i)(x-2i)=x^2+4\] and similarly
\[(x+3i)(x-3i)=x^2+9\]

Answer 5

so your last job is to write
\[f(x)=a(x^2+4)(x^2+9)\] multiply that business out, and then to find \(a\) replace x by -2, set the result equal to 104 and solve for \(a\)

Answer 6

you got this?

Answer 7

yes so far so good, then what?

Answer 8

that is it

Answer 9

o haha i did not see the last part, give me a second

Answer 10

\[f(x)=a(x^4+13 x^2+36)\] and then
\[f(-2)=104=a\left((-2)^4+13(-2)^2+36\right)\] solve for 'a'

Answer 11

does a =1?

Answer 12

i don't know i didn't compute it.
hold on

Answer 13

yes, because you get 104 for the stuff in the parentheses, so \(a=1\)

Answer 14

ok the answer i am looking for say f(x)=. so would i just put f(x)=1?