Results 1 to 4 of 4

Math Help - Polynomials

  1. #1
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7

    Polynomials

    [ \star] True or False: There exist non-zero polynomials p(x), q(x) \in \mathbb{C}[x] such that xq(x)p(x+2)=(x-1)p(x)q(x+2).
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member Bacterius's Avatar
    Joined
    Nov 2009
    From
    Wellington
    Posts
    927
    Hello NCA !
    Spoiler:

    x \times q(x) \times p(x+2)=(x-1) \times p(x) \times q(x+2)

    This is equivalent to :

    \frac{x}{x - 1} = \frac{p(x) \times q(x + 2)}{q(x) \times p(x + 2)}

    By expanding :

    x^2 - x = \frac{p(x) \times q(x + 2)}{q(x) \times p(x + 2)}

    The product of two polynomials is a polynomial. However, the division of two polynomials is never a polynomial except when the polynomial in the denominator has degree zero (which is assumed not being the case). Since x^2 - x is a polynomial, there exists no polynomials p(x), q(x) that satisfy the statement.

    Is this enough as a proof, or is it invalid ?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Jan 2009
    Posts
    715
    Given that  xq(x)p(x+2) = (x-1)p(x)q(x+2)  ~ ... ~ (1)

    By replacing  x with  x-2 ,
    we obtain :

     (x-2)q(x-2)p(x) = (x-3)p(x-2)q(x)

    then multiply it by  (1)

     xq(x)p(x+2)\cdot(x-2)q(x-2)p(x) = (x-1)p(x)q(x+2) \cdot (x-3)p(x-2)q(x)

    There are  p(x) and  q(x) on the both sides , delete them . The remaining is :


      x(x-2)p(x+2)q(x-2) = (x-1)(x-3)p(x-2)q(x+2) ~...~(2)

    Repeat this procedure again ,

    we obtain

     (x-2)(x-4)p(x)q(x-4) = (x-3)(x-5)p(x-4)q(x)

    then mutiply it by  (1) , the reference equation .

     x(x-2)(x-4)p(x+2)q(x-4) = (x-1)(x-3)(x-5)p(x-4)q(x+2) ~...~(3)

    I find that comparing with the previous equation , eg  (2)
    there is an extra linear factor on each side in  (3) and a change  p(x-2) to  p(x-4) and  q(x-2) to  q(x-4) , so I assume it is also true that :

     x(x-2)(x-4)...(x-2n)p(x+2)q(x-2n) = (x-1) (x-3)(x-5)...(x-2n-1)p(x-2n)q(x+2)

    Then at least one of them  p(x) , q(x) contains many factors , so it is not a polynomial .
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by Bacterius View Post
    Hello NCA !
    Spoiler:

    x \times q(x) \times p(x+2)=(x-1) \times p(x) \times q(x+2)

    This is equivalent to :

    \frac{x}{x - 1} = \frac{p(x) \times q(x + 2)}{q(x) \times p(x + 2)}

    By expanding :

    x^2 - x = \frac{p(x) \times q(x + 2)}{q(x) \times p(x + 2)}

    The product of two polynomials is a polynomial. However, the division of two polynomials is never a polynomial except when the polynomial in the denominator has degree zero (which is assumed not being the case). Since x^2 - x is a polynomial, there exists no polynomials p(x), q(x) that satisfy the statement.

    Is this enough as a proof, or is it invalid ?
    \frac{x^2-1}{x-1}=x+1
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 7
    Last Post: April 7th 2011, 12:38 PM
  2. GCD of polynomials in Zn[x]
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: May 18th 2010, 06:22 AM
  3. Polynomials
    Posted in the Algebra Forum
    Replies: 3
    Last Post: May 16th 2010, 06:52 AM
  4. Replies: 7
    Last Post: January 8th 2010, 03:13 AM
  5. Replies: 5
    Last Post: November 29th 2005, 03:22 PM

Search Tags


/mathhelpforum @mathhelpforum