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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
 2 years ago

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

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

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

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

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

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

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

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

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