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Algebra

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6 Jun 2008 10:47:48 IST

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Easy Polynomial Problem

None

Given a polynomial P(x) of degree 2008 and leading coefficient 1

such that P(0) = 2007; P(1) = 2006; P(2) = 2005;...;P(2007) = 0

find P(2008).

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Comments (7)

ragesh

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Joined: 4 Jun 2008

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9 Jun 2008 20:41:49 IST

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let

g(x)=P(x)-2007+x

=(x-0)(x-1)(x-2)......(x-2007) [since g(0),g(1)........,g(2007)=0]

so, P(x) =x(x-1).......(x-2007)+2007-x

now put x =2008

Mukund

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9 Jun 2008 21:05:44 IST

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Let coefficients be a₂₀₀₈ , a₂₀₀₇...... a₀

a₀ = 2007..

Summation of coefficients is 2006.. You get this from condition of P(1)

P(2008) = p(2007) + a₁ + a₂ + ... + a₂₀₀₈

= 0 + 2006 - 2007

= -1

Shreya

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9 Jun 2008 21:17:07 IST

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Ragesh , how can u say that [since g(0),g(1)........,g(2007)=0]

rates assured

Shreya

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9 Jun 2008 21:18:37 IST

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Mukund, can u tell me how the Summation of coefficients is 2006..
and also these two steps
You get this from condition of P(1)

P(2008) = p(2007) + a1 + a2 + ... + a2008

SUNDEEP ALLAMRAJU

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9 Jun 2008 22:33:13 IST

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Let us consider a function g(x) of same degree as that of P(x) defined as

g(x)=P(x)+x-2007

Then g(0),g(1),.....g(2007)=0[observe that p(x) values in general of the form P(k)=2007-k and hence P(k)+k-2007=0 for k=1 to 2007].

Therefore,(x-0),(x-1),(x-2),....(x-2007) are factors of g(x).

Hence,As degree of g(x) is 2008,g(x) is of the form k(x)(x-1)(x-2)...(x-2007).

So,P(x)=g(x)-x+2007,now,as the leading coefficient of P(x) is 1,we get,k=1.

Thus,P(x)=x(x-1)....(x-2007)-x+2007.

P(2008)=2008!-1.

gokul subramanian

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9 Jun 2008 22:42:04 IST

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i think this 1 was from AOPS? think ive done similar/same 1 from there..

Mukund

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9 Jun 2008 22:46:11 IST

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Let P (x) = a2008 x^2008 + a2007 x^2007 + ... + a0
Put x = 1
P(1) is sum of coefficients which is equal to 2006..

But I think the rest of my solution is wrong :P ...

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