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Using Intermediate Value Theorem to prove # of polynomial roots

  1. Oct 31, 2011 #1
    I've heard there's a proof out there of this, basically that (I think) you can use the intermediate value theorem to prove that an Nth-degree polynomial has no more than N roots.

    I'm not in school anymore, just an interested engineer. Does anyone know where I can find this proof or any really strong hints on how to do it myself? I've been out of it for a while and I'm rusty.
  2. jcsd
  3. Nov 1, 2011 #2


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    well the crux of any such proof is going to be that the derivative of an Nth-degree polynomial is going to be an (N-1)-th degree polynomial.

    you don't even need limits for this, you can just define a derivative formally using the power rule.

    why derivatives? because, in general, all one can say at the outset, is that a polynomial of odd degree has at least one root. to bring the IVT to bear, one has to find out how many times a polynomial p(x) can "change direction", that is, when its slope changes sign.

    for an partial example of how this becomes an induction proof:

    suppose p(x) is of degree 2, so that p(x) = ax2 + bx + c. then p'(x) = 2ax + b, (and by assumption, a ≠ 0), so p'(x) has one root (-b/(2a), in fact).

    this means that p(x) changes direction exactly once (for higher degrees it becomes "at most n-1 times" because some of the "humps" might not exist (f(x) = x3 doesn't have any, for example), so it can cross the x-axis at most twice (once going up, once going down).

    you should be able to generalize, and make this a bit more rigorous, from this little bit.
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