#1
how do you find all real and complex zeros of
10x^3 + 21x^2 - x - 6 ?
any help would be greatly appreciated
#7
look for a fairly obvious and easy to work with factor that fits like 0, 1, 2, 3/2, -1, -2, -3/2 etc and factor it out as (x - *factor*) and use long division to find the remaining quadratic and solve that for your last 2 factors.
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Last edited by Sol9989 at Aug 28, 2008,
#8
Quote by Sol9989
look for a fairly obvious factor that fits like 0, 1, 2, 3/2, -1, -2, -3/2 etc and factor it out as (x - *factor*) and use long division to find the remaining quadratic and solve that for your last 2 factors.


+1

yeah use polynomial long division.

I'm not 100% sure, but i remember something about determining factors. I think you find out whether the sum of coefficients of even degrees = the sum of coefficients of odd degrees. If it is, then (x+1) will be a factor, otherwise (x-1) will be.

Could be the other way around though, as i said, i'm still pretty sketchy on it. Google it to get more confirmation
#9
I really havent a clue what you mean by finding the zeroes


sounds dangerous though....are you trying to ****ing kill us all!?!?!?!?!
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