possible zeros of 2x^5+3x^3+5x^2-6

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possible zeros of 2x^5+3x^3+5x^2-6

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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possible zeros come from factors of 6 therefore: 1, -1, 2, -2, 3, -3 But mostly, you try 1 and -1 first
how did you get that?
isnt it factor of 6 and 2?

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Other answers:

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

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