Question

@dumbcow
State how many imaginary and real zeros the function has.
\(f(x) = x^3 - 20x^2 + 123x - 216\)
0 imaginary; 3 real
2 imaginary; 1 real
3 imaginary; 0 real
1 imaginary; 2 real

Answer 1

ahh Descartes Thm

however it may be easier to graph it
can you use a graphing calculator

Answer 2

No I don't have one. My brother stepped on it and broke it.

Answer 3

well there are plenty of online ones.... i meant for class are you allowed to graph it

Answer 4

Oh yes I am allowed to graph it.

Answer 5

http://prntscr.com/4r5tdw

Answer 6

http://www.wolframalpha.com/input/?i=plot+x%5E3%E2%88%9220x%5E2%2B123x%E2%88%92216

notice how it crosses x-axis 3 times
3 real zeros

also imaginary roots ONLY come in pairs, it is impossible to have odd num of imaginary zeros

Answer 7

Ok, and it can only have 3 zeros since it only goes up to x^3 right?

Answer 8

right all possible zeros can only add up to highest exponent

Answer 9

Ok, do you want me to open a new question for the last one or just put it here?

Answer 10

just post it here

Answer 11

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

2 - 3i is a zero of \(f(x) = x^4 - 4x^3 + 14x^2 - 4x + 13\)
I think one answer could be 2+31 since it is the conjugate but I'm not sure

Answer 12

whoops that was meant to be 2+3i not 2+31

Answer 13

you are correct, the conjugate will always be another zero

Answer 14

to find other 2 roots, you would have to use long division

or graph it and see if it crosses at an integer

Answer 15

Ok. Would I just say 2+3i is another zero of this function because it is the conjugate of 2-3i?

Answer 16

yeah because imaginary roots must come in pairs

Answer 17

Ok, thank you so much for the help!

Answer 18

yw