I need to know how to derive the zeros of the function f(x) = x^4-4x^3+30x^2-4x+29. I graphed it with a graphing calculator but it only showed as simple parabola, I need to find the complex zeros.

Question

anonymous · Answer

there are no real zeros, that is why

anonymous · Answer

that is 
$$\{i, -i, 2+5i, 2-5i\}$$

anonymous · Answer

how did you get that?

anonymous · Answer

http://www.wolframalpha.com/input/?i=x^4-4x^3%2B30x^2-4x%2B29

anonymous · Answer

perhaps you were supposed to guess i
then if i works, you know that -i must be a zero as well.  
then you know that one factor is 
$$x^2+1$$ so factor it out and get 
$$(x^2+1) (x^2-4 x+29)$$ and solve the second one

anonymous · Answer

but how was the x^2-4x+29 obtained?

anonymous · Answer

factoring. once you know that i is a zero and also -i that means you have 
$$x^4-4x^3+30x^2-4x+29=(x^2+1)\times (\text{something)}$$ so you just need to find the something. you can do it by factoring, or else by dividing

anonymous · Answer

you know "something" must look like 
$$x^2+ax+29$$ because of the x^4 and the 29, and you know a = -4 because you have to end up with -4x  and that is the only way to get it

anonymous · Answer

why do you have to end up with -4x?