Results 1 to 2 of 2

Math Help - 7th degree polynomial

  1. #1
    Junior Member
    Joined
    Jul 2008
    From
    Cyprus
    Posts
    45

    7th degree polynomial

    Hi there abit stuck with this problem, i've worked through parts and b and some of c but thats were i'm getting stuck.


    The 7th degree polynomial x^7-3x^6-7x^4+21x^3-8x+24 has a factor (x-3)

    a) Divide x^7-3x^6-7x^4+21x^3-8x+24 by x-3 and thus:

    b) express it in the form (x-3)(ax^6+bx^3+c)

    c) By putting z=x^3, find all the factors, real or complex of the 6th degree polynomial and thus:

    d) express x^7-3x^6-7x^4+21x^3-8x+24 as the product of 7 linear factors.

    I managed part a) and got x^6-7x^3-8 and part b was easy enough

    c) i substituted in Z=x^3 and got z^2-7z-8 I then used the quadratic formula and got x = 8 and x =-1

    From there i used (x-8) and (X+1) multipled them together and tried polynomial division on x^6-7x^3-8 but that didnt work

    then tried the same on x^7-3x^6-7x^4+21x^3-8x+24 and that didn't work.

    I'm at a bit of a loss now, so any help on this would be apprecciated!


    Cheers
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Nov 2008
    From
    France
    Posts
    1,458
    Hi

    So you get x^7-3x^6-7x^4+21x^3-8x+24 = (x-3)(x^3+1)(x^3-8)

    Now you just need to find the 3 cubic roots of -1 and the 3 cubic roots of 8

    For -1 this leads to
    x = r e^{i\theta}

    x^3 = r^3 e^{3i\theta} = -1 = e^{i\pi}

    r^3 = 1 then r = 1
    3\theta = \pi + 2k \pi then \theta = \frac{\pi}{3} + \frac{2k \pi}{3} with k integer


    Another way is to say that -1 is one cubic root of -1
    Therefore x^3+1 can be factorized by x+1
    x^3+1 = (x+1)(x^2-x+1)
    Then you need to solve x^2-x+1=0
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Polynomial of Degree 2
    Posted in the Differential Geometry Forum
    Replies: 5
    Last Post: January 19th 2011, 03:20 PM
  2. polynomial of degree 3
    Posted in the Algebra Forum
    Replies: 2
    Last Post: March 5th 2010, 07:28 AM
  3. low degree polynomial
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: December 14th 2009, 02:40 AM
  4. Polynomial of fifth degree
    Posted in the Calculus Forum
    Replies: 2
    Last Post: November 13th 2006, 08:25 AM
  5. 4th degree polynomial
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: September 27th 2005, 09:41 PM

Search Tags


/mathhelpforum @mathhelpforum