Results 1 to 6 of 6

Math Help - Help with polynomial

  1. #1
    Newbie
    Joined
    Apr 2008
    Posts
    7

    Help with polynomial

    How do you find the solutions to the following equation?

    x^4-x^3+x-1=0
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    Quote Originally Posted by Ericonda View Post
    How do you find the solutions to the following equation?

    x^4-x^3+x-1=0
    First, identify what kind of solution we are searching for via Des Cartes Rule of Sign.

    3 or 1 positive
    1 negative
    2 or 0 imaginary

    \frac{p}{q}=\frac{\pm 1}{\pm 1}=+1, -1

    I don't know how to format synthetic division in latex so here is the best I have at it.

    If x=1, our division looks like:

    \begin{matrix}<br />
1 & -1 & 0 & 1 & -1\\ <br />
\vdots  & 1 & 0 & 0 & 1\\<br />
1 & 0 & 0 & 1 & 0<br />
\end{matrix}

    Hence x=1 is a solution.

    (x-1)(x^3+1)

    x^3=-1
    Last edited by dwsmith; May 13th 2010 at 07:14 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Apr 2008
    Posts
    7
    the answer should be 1,-1, 1/2 + or - square root of 3/2i
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    Quote Originally Posted by Ericonda View Post
    the answer should be 1,-1, 1/2 + or - square root of 3/2i
    Ok and??

    I gave you two solutions and all the the tools to find the complex conjugate.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Apr 2008
    Posts
    7
    thank you, but i think your math skills are to hardcore for me. i just need to know how to factor it out
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    3 or 1 positive
    1 negative
    2 or 0 imaginary

    We found 1 positive solution and we know there is 1 negative solution. We have 2 choices of solutions \pm 1.

    If we take our new polynomial, x^3+1, we can do synthetic division again to obtain a quadratic polynomial.

    If x=-1, our division looks like:

    \begin{matrix}<br />
1 & 0 & 0 & 1\\ <br />
\vdots & -1 & 1 & -1\\ <br />
1 & -1 & 1 & 0<br />
\end{matrix}

    Now we have (x-1)(x+1)(x^2-x+1)

    We can now use the quadratic equation, \frac{-b\pm\sqrt{b^2-4ac}}{2a}, to solve the quadratic.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: October 23rd 2011, 06:36 AM
  2. Replies: 1
    Last Post: February 24th 2011, 06:46 PM
  3. Replies: 1
    Last Post: December 15th 2009, 07:26 AM
  4. [SOLVED] dividing polynomial by a polynomial
    Posted in the Algebra Forum
    Replies: 1
    Last Post: February 3rd 2008, 02:00 PM
  5. dividing a polynomial by a polynomial
    Posted in the Algebra Forum
    Replies: 1
    Last Post: August 2nd 2005, 12:26 AM

Search Tags


/mathhelpforum @mathhelpforum