Let P(x), Q(x), and R(x) be polynomials such that P(\(x^{5}\)) + xQ(x5) + x2R(x5) is divisible by (x4+ x3+ x2+ x +1).
Prove that P(x) is divisible by (x - 1).
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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) = f0(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:
gi(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) = g0(x^5) + xg1(x^5) + ...+ (x^4) g_4(x^5)
-P(x^5) =g0(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
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
f0(x^5) + xf1(x^5) + ...+ (x^4)f_4(x^5)
x^5 - 1 is a polynomial in x^5, thus we can let gi(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
The blog here shows 2 comments but i can see them nowhere ? Can anyone help me out ?
The blog here shows 2 comments but i can see them nowhere ?
dcklnsc