Question

Using the given zero, find one other zero of f(x). Explain the process you used to find your solution.

1 - 2i is a zero of f(x) = x4 - 2x3 + 6x2 - 2x + 5.

Answer 1

Complex zeroes always occur as conjugates
so if 1 - 2i is a zero of the function then so is 1 + 2i

Answer 2

yeah i get that but that isnt the answer is it?

Answer 3

hmm see what you mean 1 + 2i is a zero . I'm not sure exactly what they mean though. Do they want a proof that 1 + 2i is a zero?

Answer 4

im not really sure but they said they want a process that i used to find the solution and 1+2i has no process behind it so i though they wanted another zero besides the conjugate

Answer 5

well you can find another zero by dividing the function by
(x - (1 + 2i)(x - (1 - 2i)

Answer 6

see thats what i dont know how to do

Answer 7

= (x^2 + 5)

use long or synthetic division

Answer 8

how do i do that though

Answer 9

i dont know how to dived with i's in it

Answer 10

i only know know to dive if theres no imaginary munbers

Answer 11

no theres no i's its f(x) divided by x^2 + 5

Answer 12

so the whole equation divided by x^2+5 and thats it

Answer 13

how am i supposed to do that synthetically?

Answer 14

do i put -5 in the window

Answer 15

that will give you a quoitent in x^2 which you might be able to factor of there are 2 real roots As the degree of the function is 4 there are a total of 4 zeroes- 2 complex and 2 real or 4 complex.

Answer 16

im just not understanding how to get the 2 real number answers bc i dont know how to divide it

Answer 17

yea i'm trying to divide it now on paper but i cant get it to divide exactly

Answer 18

No it doesnt divide exactly There might be a mistake in the question.

Answer 19

I'm not very good at synthetic division but that wont work either - if long division hasn't

Answer 20

i figured it out just now thank you so much i got +/- i and im hoping thats right

Answer 21

i think there must be an error somewhere in the question
I've drawn the graph on my calculator and it doesn't cut the x-axis so all the zeroes must be complex

Answer 22

yeah idk i just kind of did it a different way but thank you for your help i appreciate it

Answer 23

yw

Answer 24

ah i see where i went wrong i worked out
(x - (1 +2i)(x - (1 - 2i) incorrectly
it comes to x^2 - 2x + 5

so if we divide that into f(x) we get x^2 + 1
putting x^2 + 1 = 0
x = +/- i

that is right