Polynomial Division: Show g(x) Divides f(x)

[Combinatus]

Homework Statement

Show that $$g(x) = x^3 + 1$$ divides $$f(x) = x^{9999} +1$$.

Homework Equations

The Attempt at a Solution

$$g(x)$$ can obviously be factored into the irreducible polynomials $$(x+1)(x^2 - x + 1)$$ in $$Z[x]$$, and since $$f(-1) = (-1)^{9999} + 1 = 0$$, the factor theorem gives that $$(x+1)$$ divides $$f(x)$$.

Furthermore, we get

$$x^{9999}+1 = (x^2 - x + 1) q(x) + r(x)$$

where $$r(x) = Ax+B$$ since $$deg(r(x)) < deg(x^2 - x + 1)$$ if $$r(x) \neq 0$$.

So, showing that $$A = B = 0$$ would be a good idea, which I have failed to do throughout past trials. I suspect there's an "obvious", clever trick to this, but I'm currently not seeing it.

Another approach would probably be to use $$x^{9999}+1 = (x+1)(x^2 - x + 1) q_{2}(x) + r_{2}(x)$$ where $$r_{2}(x) = Cx^2 + Dx + E$$, and so, $$x = -1$$ yields $$C - D + E = 0$$, but that hasn't gotten me anywhere either.Note: I'm assuming that I'm not supposed to use complex roots to factor $$x^2 - x + 1$$, but the problem doesn't specify that such an assumption is necessary.

 

[Mark44]

Homework Statement

Show that $$g(x) = x^3 + 1$$ divides $$f(x) = x^{9999} +1$$.

Homework Equations

The Attempt at a Solution

$$g(x)$$ can obviously be factored into the irreducible polynomials $$(x+1)(x^2 - x + 1)$$ in $$Z[x]$$, and since $$f(-1) = (-1)^{9999} + 1 = 0$$, the factor theorem gives that $$(x+1)$$ divides $$f(x)$$.

Furthermore, we get

$$x^{9999}+1 = (x^2 - x + 1) q(x) + r(x)$$

where $$r(x) = Ax+B$$ since $$deg(r(x)) < deg(x^2 - x + 1)$$ if $$r(x) \neq 0$$.

So, showing that $$A = B = 0$$ would be a good idea, which I have failed to do throughout past trials. I suspect there's an "obvious", clever trick to this, but I'm currently not seeing it.

Another approach would probably be to use $$x^{9999}+1 = (x+1)(x^2 - x + 1) q_{2}(x) + r_{2}(x)$$ where $$r_{2}(x) = Cx^2 + Dx + E$$, and so, $$x = -1$$ yields $$C - D + E = 0$$, but that hasn't gotten me anywhere either.


Note: I'm assuming that I'm not supposed to use complex roots to factor $$x^2 - x + 1$$, but the problem doesn't specify that such an assumption is necessary.

You've shown that x + 1 is a factor by showing that f(-1) = 0. The other two factors of x^3 + 1 are x = 1/2 +/- i sqrt(3)/2. If you write these in polar form it's pretty easy to raise them to the 9999th power, and thus show that f(1/2 +/- i sqrt(3)/2) = 0.

 

[Combinatus]

You've shown that x + 1 is a factor by showing that f(-1) = 0. The other two factors of x^3 + 1 are x = 1/2 +/- i sqrt(3)/2. If you write these in polar form it's pretty easy to raise them to the 9999th power, and thus show that f(1/2 +/- i sqrt(3)/2) = 0.

Ugh, yay for assumptions! Thank you!