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Skaematik Group TitleBest ResponseYou've already chosen the best response.0
possible zeros come from factors of 6 therefore: 1, 1, 2, 2, 3, 3 But mostly, you try 1 and 1 first
 one year ago

magepker728 Group TitleBest ResponseYou've already chosen the best response.0
how did you get that?
 one year ago

magepker728 Group TitleBest ResponseYou've already chosen the best response.0
isnt it factor of 6 and 2?
 one year ago

Skaematik Group TitleBest ResponseYou've already chosen the best response.0
That is a fact. Possible zeroes come from factors of the coefficient independent of x
 one year ago

Skaematik Group TitleBest ResponseYou've already chosen the best response.0
Use the remainder theory to find the zeroes for the polynomial
 one year ago

Skaematik Group TitleBest ResponseYou've already chosen the best response.0
if P(x) = 0, x is a zero.
 one year ago

magepker728 Group TitleBest ResponseYou've already chosen the best response.0
well think you can find it or me for now?
 one year ago

Skaematik Group TitleBest ResponseYou've already chosen the best response.0
Find it yourself I just told you how to find it
 one year ago

magepker728 Group TitleBest ResponseYou've already chosen the best response.0
lol im on a test bro..
 one year ago

Yahoo! Group TitleBest ResponseYou've already chosen the best response.0
2x^5+3x^3+5x^26 = 0 No of possible for Real Positive roots is 1 no of possible for Real ve roots is 2 or 0
 one year ago

cinar Group TitleBest ResponseYou've already chosen the best response.0
since f(0)<0 and f(1)>0 one root is between 0 and 1
 one year ago

cinar Group TitleBest ResponseYou've already chosen the best response.0
Bolzano theorem
 one year ago

magepker728 Group TitleBest ResponseYou've already chosen the best response.0
i have to use the rational zeros theorem
 one year ago

cinar Group TitleBest ResponseYou've already chosen the best response.0
root is 0.83271 it is not integer, so it is not easy
 one year ago
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