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Complex zeroes always occur as conjugates so if 1 - 2i is a zero of the function then so is 1 + 2i
yeah i get that but that isnt the answer is it?
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?
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
well you can find another zero by dividing the function by (x - (1 + 2i)(x - (1 - 2i)
see thats what i dont know how to do
= (x^2 + 5) use long or synthetic division
how do i do that though
i dont know how to dived with i's in it
i only know know to dive if theres no imaginary munbers
no theres no i's its f(x) divided by x^2 + 5
so the whole equation divided by x^2+5 and thats it
how am i supposed to do that synthetically?
do i put -5 in the window
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.
im just not understanding how to get the 2 real number answers bc i dont know how to divide it
yea i'm trying to divide it now on paper but i cant get it to divide exactly
No it doesnt divide exactly There might be a mistake in the question.
I'm not very good at synthetic division but that wont work either - if long division hasn't
i figured it out just now thank you so much i got +/- i and im hoping thats right
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
yeah idk i just kind of did it a different way but thank you for your help i appreciate it
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