Question

find the polynomial function of degree 4 with zeros 2i and 1-i and a constant term of -24

Answer 1

do you have any thoughts?

Answer 2

your polynomial has complex roots
if \(z=a+ib\) is a zero then \(\bar{z}=a-ib\) also is a zero

Answer 3

Other than the fact that we only have two zeros, so at least one of them has multiplicity >1 I don't know where to start.

Answer 4

just told u

Answer 5

u HAVE (can deduce) the 4 zeros u need

Answer 6

so b is 2 and a is 1?

Answer 7

no in one case u have \(z=2i\) and in the other \(z=1-i\)

Answer 8

so then \[(0+2i)(0-2i)\] and \[(1-i)(1+i)\]

Answer 9

yes your polynomial should be
\[ \large f(x)=\alpha(1-i)(1+i)(2i)(-2i) \]

Answer 10

what do you use the \(\alpha\) for?

Answer 11

thanks! is alpha the constant -24?

Answer 12

I'm not sure, is alpha -24?

Answer 13

do the multiplication. u'll figure it out

Answer 14

If I factor out 2 and -2 it seems like 24 should be -6, but I get confused with complex numbers

Answer 15

I mean alpha should be -6

Answer 16

sorry made a mistake
\[ \large f(x)=\alpha(x-1-i)(x-1+i)(x-2i)(x+2i) \]

Answer 17

lets see
\[ \large f(x)=\alpha(x^2-2x+(1-i)(1+i))(x^2+4) \]
you can go on

Answer 18

so then alpha should be -8?

Answer 19

give me a second

Answer 20

if u dont take into account the \(\alpha\) you get
\[ \large f(x)=\alpha(x^4-2x^3+6x^2-8x+8) \]

Answer 21

so then its -3