Results 1 to 2 of 2

Math Help - Descarte's Theorem

  1. #1
    Junior Member
    Joined
    Feb 2007
    Posts
    33

    Descarte's Theorem

    Using Descartes theorem, find the possible numbers of positive, negative, and non-real complex numbers:

    P(x) = x^3 - 5x^2 + 3x + 7
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by Trentt View Post
    Using Descartes theorem, find the possible numbers of positive, negative, and non-real complex numbers:

    P(x) = x^3 - 5x^2 + 3x + 7
    Descartes' rule of signs says that the number of positive roots is equal at
    most to the number of sign changes in the coefficients of a polynomial, and
    that the number of actual roots can only be a multiple of 2 less than this
    number.

    In this case there are 2 sign change, so there are 2 or 0 positive real roots

    The possible number of negative roots can be investigates by considering
    the sign changes in P(-x) = -x^3 - 5x^2 - 3x +7, so there is at most 1
    negative root (as there is only one sign change). But there must be exactly
    one such root as 1-2, 1-3, .. cannot be the number of roots.

    So we conclude that there is 1 negative root, and there are either 0 or 2
    positive roots, and if there are 0 positive there are 2 non-real complex roots,
    and if 2 positive roots there are 0 non-real complex roots.

    RonL
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 4
    Last Post: January 10th 2011, 09:51 AM
  2. Replies: 3
    Last Post: May 14th 2010, 11:04 PM
  3. Replies: 2
    Last Post: April 3rd 2010, 05:41 PM
  4. Replies: 0
    Last Post: November 13th 2009, 06:41 AM
  5. Please solve using Descarte's Rule of Signs
    Posted in the Pre-Calculus Forum
    Replies: 5
    Last Post: October 30th 2007, 06:07 AM

Search Tags


/mathhelpforum @mathhelpforum