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[SPAM] Re: [obm-l] Re: [obm-l] Problema com polinômios
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- Subject: [SPAM] Re: [obm-l] Re: [obm-l] Problema com polinômios
- From: "Igor Battazza" <battazza@xxxxxxxxx>
- Date: Mon, 14 Jan 2008 20:19:33 -0200
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Essa lista é mesmo fantastica! Obrigado a todos mesmo!
Com relação ao problema proposto pelo prof. Shine:
>
> Note que se fossem 3 inteiros a,b,c no lugar de
> a,b,c,d, seria possível construir P(x). De fato,
> fazendo as mesmas contas (sem o d, claro) obtemos
> 3 = Q(k) = (k-a)(k-b)(k-c)R(x)
> e podemos tomar k-a = -3, k-b = -1 e k-c = 1. Tomando
> k=0, temos a = 3, b = 1 e c = -1. Tomando ainda R(x) =
> 1, obtemos
> Q(x) = (x-3)(x-1)(x+1) = x^3 - 3x^2 - x + 3
> ou, mudando para P(x),
> P(x) = Q(x) + 5 = x^3 - 3x^2 - x + 8.
>
> Temos P(3) = P(1) = P(-1) = 5 (verifique!) e P(0) = 8.
>
> Você conseguiria encontrar *todos* os polinômios P(x)
> desse novo problema?
>
> []'s
> Shine
>
Se eu entendi bem, o problema seria encontrar todos os polinômios P(x)
onde existem inteiros distintos a, b e c tal que P(a) = P(b) = P(c) =
5?
Poderia supor a existencia de inteiros k, t tal que P(k) = t. (I)
Usando a mesma idéia do problema anterior: Seja Q(x) = P(x) - 5, onde
a, b e c são raizes de Q(x). Portanto Q(x) = (x-a)(x-b)(x-c)R(x), e de
(I) teriamos:
Q(k) = P(k) - 5 = t - 5 = (k-a)(k-b)(k-c)R(x)
Agora eu engasguei... Pensei em usar tau(t-5), onde poderia obter o
numero de divisores (positivos), não sei se ajudaria em algo...
(Desculpem se estou dizendo asneiras)
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