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    Find (without calculus) a fifth degree polynomial p(x) such that p(x) + 1 is divisible by (x - 1)^3 and p(x) - 1 is divisible by (x + 1)^3.

    In one solution I found, the following properties were used (given without proof) but I am not sure why those properties are so obvious.

    1) If p(x) + 1 is divisible by (x-1)^3, then p(-x) -1 is also divisible by (x-1)^3.
    2) If p(x) - 1 is divisible by (x + 1)^3 then p(-x) + 1 is also divisible by (x+1)^3.

    Thanks in advance.
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    (Original post by tangotangopapa2)
    Find (without calculus) a fifth degree polynomial p(x) such that p(x) + 1 is divisible by (x - 1)^3 and p(x) - 1 is divisible by (x + 1)^3.

    In one solution I found, the following properties were used (given without proof) but I am not sure why those properties are so obvious.

    1) If p(x) + 1 is divisible by (x-1)^3, then p(-x) -1 is also divisible by (x-1)^3.
    2) If p(x) - 1 is divisible by (x + 1)^3 then p(-x) + 1 is also divisible by (x+1)^3.
    Since p(x) + 1 = k(x-1)^3 \Rightarrow p(x) = k(x-1)^3 - 1 then p(-x) - 1 = k(-x-1)^3 + 1 -1 = -k(x+1)^3, etc...
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    (Original post by Zacken)
    Since p(x) + 1 = k(x-1)^3 \Rightarrow p(x) = k(x-1)^3 - 1 then p(-x) - 1 = k(-x-1)^3 + 1 -1 = -k(x+1)^3, etc...
    Thank you for the reply. As p(x) is a fifth degree polynomial, k should be a second degree polynomial not a constant. If you replace x by -x, k no longer remains same.

    Would you mind telling me flaw(if there is) in above argument?
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    (Original post by tangotangopapa2)
    Thank you for the reply. As p(x) is a fifth degree polynomial, k should be a second degree polynomial not a constant. If you replace x by -x, k no longer remains same.

    Would you mind telling me flaw(if there is) in above argument?
    My argument was wrong because I failed to add correctly, but your point is incorrect. We don't care about k, it's just a factor.

    Anywho: corrected:

    Since p(x) - 1 = q(x)(x+1)^3. Then p(-x) -1 = q(-x) (-x+1)^3 = -q(-x) (x-1)^3.

    In part, sure, q changed to q(-), but it's still just a floating factor and of no consequence. The end result is that p(-) - 1 is made up of two factors -q(-x) and (x-1)^3 and hence divisible by (x-1)^3.
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    A shame you can't use calculus. Makes it really really easy.
    Anywho, the standard writing p(x)=ax^5+ etc
    Solve for you coefficients the -x makes sure when solving simultaneous eqs adding will cancel some coefficients and recall values of p(1) and p(-1) and a few others and you should be done.


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    (Original post by physicsmaths)
    A shame you can't use calculus. Makes it really really easy.
    Anywho, the standard writing p(x)=ax^5+ etc
    Solve for you coefficients the -x makes sure when solving simultaneous eqs adding will cancel some coefficients and recall values of p(1) and p(-1) and a few others and you should be done.


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    It is British Mathematics Olympiad question and it is asked to solve without Calculus. I am comfortable with Calculus.

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    (Original post by tangotangopapa2)
    It is British Mathematics Olympiad question and it is asked to solve without Calculus. I am comfortable with Calculus.

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    What year?
    I did all the questions from late 90s to now.


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    (Original post by physicsmaths)
    What year?
    I did all the questions from late 90s to now.


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    I am sorry, I thought it was question from past. Just checked it out. It is from one of my practice books : http://www.cs.elte.hu/~nagyzoli/tematic.pdf

    Question 1 from polynomial section.
 
 
 
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