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

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

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

Skaematik
 2 years ago
Best ResponseYou've already chosen the best response.0That is a fact. Possible zeroes come from factors of the coefficient independent of x

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

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

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

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

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

Yahoo!
 2 years ago
Best ResponseYou've already chosen the best response.02x^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

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

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

cinar
 2 years ago
Best ResponseYou've already chosen the best response.0root is 0.83271 it is not integer, so it is not easy
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