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Roots of a nth degree polynomial

  1. Nov 13, 2013 #1
    why does a nth degree polynomial has atleast one root and a maximum of n root...?

    In my book it's given, it's the fundamental theorem of algebra.

    Is there a proof....?

    Thank's for help. (In advance)
  2. jcsd
  3. Nov 13, 2013 #2
    Roots over what? A polynomial with real coefficients does not necessarily have any real roots, though it does have n roots over the complex numbers. The wikipedia article on "The fundamental theorem of algebra" contains several proofs.
  4. Nov 13, 2013 #3
    It's not specified, i guess it over real number. I saw wikipedia but the proof are difficult .
  5. Nov 13, 2013 #4


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    Sadly, many proofs are difficult. Some are even book length.
  6. Nov 14, 2013 #5


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    The "fundamental theorem of algebra" is normally stated as "every polynomial equation has at least one root in the complex numbers." Since a root, a, implies "z- a" is a factor, we can reduce to another equation of degree n-1, which has a root, then to a polynomial of degree n-2, which has a root, etc. until we are reduce to a linear polynomial. What we might call "the extended fundamental of algebra" says that any nthe degree polynomial has n roots where we are counting "multiple roots". That is, [itex]z^3- 3z^2+ 3z- 1= 0[/itex], [itex](z- 1)^3= 0[/itex] has three roots, all of them equal to "1".

    Yes, there are a number of proofs. Two distinctly different proofs, one using very basic properties of algebra and fairly lengthy, the other much more sophisticated and shorter.

    The simpler, but longer, proof can be found on Wikipedia:
    which I found by googling "fundamental theorem of algebra".
  7. Nov 15, 2013 #6


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    Unless your coefficients are [strike]p-adics[/strike] :tongue:

    Edit: Well technically ##C_p## is isomorphic to ##\mathbb{C}## but only through AoC.
    Last edited: Nov 15, 2013
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