Results 1 to 3 of 3
Like Tree1Thanks
  • 1 Post By Soroban

Math Help - Factoring cubic polynomials

  1. #1
    Member
    Joined
    May 2009
    Posts
    109

    Factoring cubic polynomials

    I do not understand how we know a cubic polynomial such as

    z^3 -3z^2 + 4z - 2 = 0

    can be factored as

    (z - 1)(z^2 - 2z +2) = 0

    I can, of course, work it backwards but is this actually some kind of identity? and if so what is its derivation?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Jun 2009
    Posts
    660
    Thanks
    133

    Re: Factoring cubic polynomials

    At this sort of level, it's pretty much a case of trial and error.

    Substitute values for z and hope that one of them gets you a zero value for the polynomial.

    Start off by trying z = 0,1,2,-1,-2,3 etc. For this particular example you get a hit with z = 1 in which case z - 1 is a factor and you can now divide out to determine the quadratic.

    Notice though that if for two of your trial values the polynomial changes sign then there has to be a zero between those two values. So for your example z = 0 and z = 2 gets you polynomial values of -2 and +2 in which case there has to be a value of z between 0 and 2 making the poynomial zero. (In general this need not be an integer value.)

    If you get no joy from this method you might consider sketching a graph or graphs (usually graphs is better).
    For this example rewrite the equation as z^{3}-3z^{2}=2-4z and sketch the graphs of y = z^{3}-3z^{2} and y = 2 -4z. The values of z at which the graphs intersect will be the roots of the equation.
    For this example this will tell you that there is a root somewhere between 1/2 and 3.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,735
    Thanks
    642

    Re: Factoring cubic polynomials

    Hello, alyosha2!

    I do not understand how we know a cubic polynomial such as: . x^3 -3x^2 + 4x - 2 \:=\: 0

    can be factored as: . (x - 1)(x^2 - 2z +2) \:=\:0

    We are given a polynomial equation: . P(x) \:=\:0

    We are expected to know two theorems.

    [1] If P(a) = 0, then x = a is a root of the equation
    . . .and x-a is a factor of P(x).

    [2] If P(x) has a rational root, it is of the form: x \,=\,\frac{n}{d}
    . . .where n is a factor of the constant term
    . . .and d is a factor of the leading coefficient.


    Our polynomial is: . x^3 - 3x^2 + 4x - 2
    Its constant term is 2; its factors are: \pm1,\:\pm2
    Its leading coefficient is 1; its factor are: \pm1

    Hence, the only possible rational roots are: . \pm1,\:\pm2

    So, we try each of them until one of them makes P(x) equal to zero.


    We find x = 1 is a zero of the polynomial.
    . . Hence, (x-1) is a factor.

    Using long division, we have: . x^3 - 3x^2 + 4x - 23 \:=\:(x-1)(x^2-2x+2)
    Thanks from MarkFL
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Factoring a cubic
    Posted in the Algebra Forum
    Replies: 1
    Last Post: January 30th 2011, 01:22 PM
  2. Cubic polynomials
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 18th 2009, 05:22 AM
  3. Factoring cubic and further polynomials
    Posted in the Algebra Forum
    Replies: 5
    Last Post: August 31st 2009, 02:16 PM
  4. cubic polynomials
    Posted in the Algebra Forum
    Replies: 2
    Last Post: May 3rd 2008, 06:01 PM
  5. Replies: 4
    Last Post: May 2nd 2007, 08:41 AM

Search Tags


/mathhelpforum @mathhelpforum