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OpenStudy (anonymous):
possible zeros of
2x^5+3x^3+5x^2-6
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OpenStudy (anonymous):
possible zeros come from factors of 6
therefore: 1, -1, 2, -2, 3, -3
But mostly, you try 1 and -1 first
OpenStudy (anonymous):
how did you get that?
OpenStudy (anonymous):
isnt it factor of 6 and 2?
OpenStudy (anonymous):
That is a fact. Possible zeroes come from factors of the coefficient independent of x
OpenStudy (anonymous):
Use the remainder theory to find the zeroes for the polynomial
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OpenStudy (anonymous):
if P(x) = 0, x is a zero.
OpenStudy (anonymous):
well think you can find it or me for now?
OpenStudy (anonymous):
Find it yourself I just told you how to find it
OpenStudy (anonymous):
lol im on a test bro..
OpenStudy (anonymous):
2x^5+3x^3+5x^2-6 = 0
No of possible for Real Positive roots is 1
no of possible for Real -ve roots is 2 or 0
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OpenStudy (anonymous):
since f(0)<0 and f(1)>0 one root is between 0 and 1
OpenStudy (anonymous):
Bolzano theorem
OpenStudy (anonymous):
i have to use the rational zeros theorem
OpenStudy (anonymous):
root is 0.83271 it is not integer, so it is not easy
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