Waste less time on Facebook — follow Brilliant.
×

Polynomial problem

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

Note by Nishant Sharma
4 years, 3 months ago

No vote yet
1 vote

Comments

Sort by:

Top Newest

dcklnsc Nishant Sharma · 4 years, 3 months ago

Log in to reply

The blog here shows 2 comments but i can see them nowhere ? Nishant Sharma · 4 years, 3 months ago

Log in to reply

The blog here shows 2 comments but i can see them nowhere ? Can anyone help me out ? Nishant Sharma · 4 years, 3 months ago

Log in to reply

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 Gabriel Wong · 4 years, 3 months ago

Log in to reply

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 Gabriel Wong · 4 years, 3 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...