Results 1 to 3 of 3

Math Help - General Question about Cubic Polynomials

  1. #1
    Super Member
    Joined
    Jun 2009
    From
    United States
    Posts
    676
    Thanks
    19

    General Question about Cubic Polynomials

    I've been studying methods for finding the roots of cubic functions. The general formulae seem overwhelmingly difficult to remember. I know it customary for students to memorize the quadratic formula, but what about the formula seen here: Cubic Formula -- from Wolfram MathWorld

    ?

    LOL, how could anybody possilby do something like that on a test? The cubic formula seems impossible to remember, so I just made up a different method. Say I have a cubic equation x^3-6x^2+21x-26=0.

    f(x)=x^3-6x^2+21x-26

    f'(x)=3x^2-12x+21

    f''(x)=6x-12

    If you look at the graphs of cubic functions, there always seems to be a point of inflection at the x-intercept. Is this true in general? If so, then f''(x)=0 -> x=2 gives the root f(2)=0, So now I proceed with synthetic division. Is this method something I can rely on? Are there always roots at the inflection points of cubic functions?

    Generally,

    f(x)=ax^3+bx^2+cx+d

    f'(x)=3ax^2+2bx

    f''(x)=6ax+2b

    f''(x)=0 -> x=-\frac{b}{3a}

    Notice that \frac{-(-6)}{3}=2

    I'm not good at proving things in math, so I don't know if I can claim that this works in general, and that all cubic functions will have at least one root of the form (x+\frac{b}{3a})
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,894
    Thanks
    691
    both of the following cubics have an inflection point at x = 2

    f(x) = x^3 - 6x^2 + 3x - 18

    f(x) = x^3 - 6x^2 - 5x + 30

    do either have a root at x = 2 ?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Jun 2009
    From
    United States
    Posts
    676
    Thanks
    19
    Quote Originally Posted by skeeter View Post
    both of the following cubics have an inflection point at x = 2

    f(x) = x^3 - 6x^2 + 3x - 18

    f(x) = x^3 - 6x^2 - 5x + 30

    do either have a root at x = 2 ?

    ok, so my method works mererly as a test. Set f''(x)=0, maybe you'll find a root. The 0=x^3-6x^2+3x-18 is easy to solve by factoring, so this one wouldn't be a problem. Other cubic function are more difficult to deal with, so I'm trying to explore simplified methods of finding the roots.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Cubic polynomials
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 18th 2009, 05:22 AM
  2. Solving cubic polynomials
    Posted in the Algebra Forum
    Replies: 9
    Last Post: October 26th 2009, 11:55 PM
  3. Factoring cubic and further polynomials
    Posted in the Algebra Forum
    Replies: 5
    Last Post: August 31st 2009, 02:16 PM
  4. Cubic polynomials- finding unknowns
    Posted in the Algebra Forum
    Replies: 3
    Last Post: June 15th 2009, 03:55 AM
  5. cubic polynomials
    Posted in the Algebra Forum
    Replies: 2
    Last Post: May 3rd 2008, 06:01 PM

Search Tags


/mathhelpforum @mathhelpforum