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 Sextic Settlement (Posted on 2015-05-13)
P is a polynomial of degree 6. M and N are two real numbers with 0 < M < N. Given that:
1. P(M) = P(-M), and:
2. P(N) = P(-N)
3. P’(0) = 0
Does the relationship P(x) = P(-x) hold for all nonzero real values of x?
If so, prove it.
If not, provide a counterexample.

 No Solution Yet Submitted by K Sengupta Rating: 3.5000 (2 votes)

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 Counter Counter example (spoiler) | Comment 2 of 6 |
Yes, it is true, and it holds for all values of x (zero or non-zero).

We only need to consider the odd powered terms.

Let Q(x) = bx^5 + dx^3 + fx

From 3, we know that f = 0

So Q(x) = bx^5 + dx^3

We have bM^5 + dM^3 = b(-M)^5 + d(-M)^3
If b <> 0 then d/b = -M^2

Similarly,
If b <> 0 then d/b = -N^2

This is a contradiction, so b = 0, and so d = 0.

So P(x) = ax^6 + cx^4 + ex^2 + g
So P(x) = -P(x) for all x

 Posted by Steve Herman on 2015-05-13 08:35:31

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