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1. 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.

2. (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 then , etc...
3. (Original post by Zacken)
Since then , 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?
4. (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 , it's just a factor.

Anywho: corrected:

Since . Then .

In part, sure, changed to , but it's still just a floating factor and of no consequence. The end result is that is made up of two factors and and hence divisible by .
5. 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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6. (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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7. (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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8. (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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Updated: August 5, 2016
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