anonymous
  • anonymous
possible zeros of 2x^5+3x^3+5x^2-6
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
possible zeros come from factors of 6 therefore: 1, -1, 2, -2, 3, -3 But mostly, you try 1 and -1 first
anonymous
  • anonymous
how did you get that?
anonymous
  • anonymous
isnt it factor of 6 and 2?

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anonymous
  • anonymous
That is a fact. Possible zeroes come from factors of the coefficient independent of x
anonymous
  • anonymous
Use the remainder theory to find the zeroes for the polynomial
anonymous
  • anonymous
if P(x) = 0, x is a zero.
anonymous
  • anonymous
well think you can find it or me for now?
anonymous
  • anonymous
Find it yourself I just told you how to find it
anonymous
  • anonymous
lol im on a test bro..
anonymous
  • 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
anonymous
  • anonymous
since f(0)<0 and f(1)>0 one root is between 0 and 1
anonymous
  • anonymous
Bolzano theorem
anonymous
  • anonymous
i have to use the rational zeros theorem
anonymous
  • anonymous
root is 0.83271 it is not integer, so it is not easy

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