**P(x)**, **Q(x)**, and **R(x)** be polynomials such that **P(\(x^{5}\))** + x**Q(\(x^{5}\))** + \(x^{2}\)**R(\(x^{5}\))** is divisible by (\(x^{4}\:\)+ \(x^{3}\:\)+ \(x^{2}\:\)+ x +1).
Prove that **P(x)** is divisible by (x - 1).

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The blog here shows 2 comments but i can see them nowhere ?

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The blog here shows 2 comments but i can see them nowhere ? Can anyone help me out ?

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Weird...my previous post didnt show up.

(x-1) { P(x^5) + xQ(x^5) + (x^2) R(x^5) } = f(x) (x^5 - 1)

decompose f(x) into polynomials

f

0(x^5) + xf1(x^5) + ...+ (x^4)f_4(x^5)x^5 - 1 is a polynomial in x^5, thus we can let g

i(x^5) = (x^5-1)fi(x^5)Comparing coefficients of (x-1) { P(x^5) + xQ(x^5) + (x^2) R(x^5) } = f(x) (x^5 - 1),

P(x^5) = -g_0(x^5)

Since x^5 covers all reals, replace x^5 with y. we now have:

P(y) = -(y-1)f_0(y)

which is the desired result

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P(x^5) + xQ(x^5) + (x^2)R(x^5) = f(x)(x^4+x^3+x^2+x+1) for some polynomial f(x)

multiplying both sides by (x-1)

(x-1) { P(x^5) + xQ(x^5) + (x^2)R(x^5) } = f(x) (x^5 -x^0)

(x^3)R(x^5) + (x^2){ Q(x^5) - R(x^5) } + x { P(x^5) - Q(x^5) } - P(x^5) = f(x) (x^5 - x^0)

let f(x) = f

0(x^5) + xf1(x^5) + ...+ (x^4) f4(x^5) for polynomials f0,f1...,f4. This decomposition is unique.g(x) = f(x)(x^5-x^0) can be decomposed in the same manner. Because x^5 - x^0 is a polynomial in x^5, we have:

g

i(x^5) = (x^5-x^0)fi(x^5)Comparing coefficients,

(x^3)R(x^5) + (x^2){ Q(x^5) - R(x^5) } + x { P(x^5) - Q(x^5) } - P(x^5) = g

0(x^5) + xg1(x^5) + ...+ (x^4) g_4(x^5)-P(x^5) =g

0(x^5) = (x^5-1)f0(x^5)x^5 covers all reals, letting x^5 = y

P(y) = -(y-1)f_0(y)

which yields the desired result

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