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  1. #1
    rcs
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    Find the polynomials p (x)

    please help me on this. thank you so much

    Find all polynomials p(x) where xp(x -1) = (x-5)p(x) and p(6) = 5 factorials
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    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by rcs View Post
    Find all polynomials p(x) where xp(x -1) = (x-5)p(x) and p(6) = 5 factorials
    Hint :

    If p(x) satisfies xp(x-1)=(x-5)p(x) then,

    For x=0 , 0=-5p(0) that is, p(0)=0 .

    For x=1 , 1p(0)=-4p(1) that is, p(1)=0 .

    ...

    x=0,1,2,3,4 are roots of p(x)


    Fernando Revilla
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    Didn't we just recently have this?

    Here's how I would do it (rather different from the way FernandoRevilla did it). Taking x= 5, 5p(4)= 0 so that p(4)= 0. Since p is a polynomial, that means that p has a factor of x- 4: we can write p(x)= (x- 4)q(x) for some polynomial q.

    Putting that into the given equation, xp(x- 1)= x((x- 1)- 4)q(x-1)= x(x- 5)q(x-1)= (x- 5)(x- 4)q(x). Since this is true for all x, we can cancel the "x- 5" terms an see that q satisfies
    xq(x- 1)= (x- 4)q(x).

    Talking x= 4, we have 4q(3)= 0 so that q(3)= 0 and we can write q(x)= (x- 3)r(x) for some polynomial r.

    Now xq(x-1)= x(x-1- 3)r(x-1)= (x- 4)(x- 3)r(x). Since this is true for all x, we cancel the "x-4" terms and have

    xr(x-1)= (x- 3)r(x). Taking x= 3, we have 3r(2)= 0 so that r(x)= (x- 2)s(x) for some polynomial s.

    Get the point? Keep doing that until it terminates (it will).
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    Quote Originally Posted by HallsofIvy View Post
    Didn't we just recently have this?
    Yes.
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  5. #5
    rcs
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    answer from the book without solution

    Find the polynomials p (x)-kkkk.jpg


    hello everybody Dear Mathematicians... i saw the answer from where i got that problem i posted above... this is the answer but i don't know how they reach to this
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    MHF Contributor FernandoRevilla's Avatar
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    Talking

    Quote Originally Posted by rcs View Post
    Click image for larger version. 

Name:	kkkk.jpg 
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    hello everybody Dear Mathematicians... i saw the answer from where i got that problem i posted above... this is the answer but i don't know how they reach to this
    We have proved that x=0,1,2,3,4 are roots of p(x) so, all the polynomials satisfying

    xp(x-1)=(x-5)p(x)\quad[*]

    have the form:

    p(x)=x(x-1)(x-2)(x-3)(x-4)q(x)

    On the other hand, [*] implies:

    x(x-1)(x-2)(x-3)(x-4)(x-5)q(x-1)=

    (x-5)x(x-1)(x-2)(x-3)(x-4)q(x)

    that is, q(x)=q(x-1) and necessarily q(x)=a constant ( Why ?).

    So, all polynomials p(x) satisfying [*] are:

    p(x)=ax(x-1)(x-2)(x-3)(x-4)

    and the condition p(6)=5! implies a=1/6 .

    Fernando Revilla
    Last edited by FernandoRevilla; Jan 15th 2011 at 07:26 AM.
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