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Thread: Proof

  1. #1 Proof 
    Forum Ph.D. Heinsbergrelatz's Avatar
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    the polynomial where a, b,c,d are all integers. If p and q are two relatively prime integers, show that if is a factor P(x), then p is a factor d, and q is a factor of a.

    please, i need some explanation, thank you.


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  3. #2 Re: Proof 
    . DrRocket's Avatar
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    Quote Originally Posted by Heinsbergrelatz
    the polynomial where a, b,c,d are all integers. If p and q are two relatively prime integers, show that if is a factor P(x), then p is a factor d, and q is a factor of a.

    please, i need some explanation, thank you.
    What explanation do you seek ?

    The assertion is quite obvious.


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  4. #3  
    Forum Ph.D. Leszek Luchowski's Avatar
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    If is a factor of , and is a third-degree polynomial, it means that there exists a second-degree polynomial such that .

    Let us write S(x) as the polynomial it's supposed to be:



    Now you do the rest.
    Leszek. Pronounced [LEH-sheck]. The wondering Slav.
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  5. #4  
    Forum Ph.D. Heinsbergrelatz's Avatar
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    can somebody complete the proof please? im not trying to get it the easy way, i just want a proper proof from someone more experienced...


    thanks
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  6. #5  
    Forum Ph.D. Leszek Luchowski's Avatar
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    Heinsberg, I gave you about 3/4 of a complete proof. You just have to show that you are putting some effort of your own into it.

    Do what you can, even if it is erroneous, and I'll be happy to point out any mistakes, or suggest the next step if you get stuck. It will cost me more time and brainpower than it would to just finish what I began in my previous post, but it will give me the rare luxury of not feeling that I'm being fooled into doing somebody's work for them.
    Leszek. Pronounced [LEH-sheck]. The wondering Slav.
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  7. #6  
    Forum Ph.D. Heinsbergrelatz's Avatar
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    Although it might seem completely absurd, here it goes..

    When q is a factor of a, there must be some " " to equal a.same goes for when p is a factor of d.

    now of we expand the second degree polynomial with the factor it goes;



    thereby; and

    can you correct any mistake you see?? thanks
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  8. #7  
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    Quote Originally Posted by Heinsbergrelatz
    Although it might seem completely absurd, here it goes..

    When q is a factor of a, there must be some " " to equal a.same goes for when p is a factor of d.

    now of we expand the second degree polynomial with the factor it goes;



    thereby; and

    can you correct any mistake you see?? thanks
    That's beautiful.
    Wise men speak because they have something to say; Fools, because they have to say something.
    -Plato

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  9. #8  
    Forum Ph.D. Heinsbergrelatz's Avatar
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    That's beautiful.
    sarcasm?? or you really mean it?? im not being suspicious but i have seen many sarcastic compliments in this forum.. But if its not a sarcasm, thanks
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  10. #9  
    . DrRocket's Avatar
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    Quote Originally Posted by Heinsbergrelatz
    That's beautiful.
    sarcasm?? or you really mean it?? im not being suspicious but i have seen many sarcastic compliments in this forum.. But if its not a sarcasm, thanks
    Your proof is correct, and it really is that simple.
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  11. #10  
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    Quote Originally Posted by Heinsbergrelatz
    That's beautiful.
    sarcasm?? or you really mean it?? im not being suspicious but i have seen many sarcastic compliments in this forum.. But if its not a sarcasm, thanks
    No sarcasm. sincere statement, which was reinforced by Dr. Rocket
    Wise men speak because they have something to say; Fools, because they have to say something.
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  12. #11  
    Forum Ph.D. Leszek Luchowski's Avatar
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    Heinsberg, you seem reluctant to believe this, but it's a fact: you've done it, man!

    I'm genuinely glad. Keep up the good work.
    Leszek. Pronounced [LEH-sheck]. The wondering Slav.
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  13. #12  
    Forum Ph.D. Heinsbergrelatz's Avatar
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    Thanks Everyone for the confirmation :-D

    I'm genuinely glad. Keep up the good work.
    thanks
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