Factoring $X^{16}+X$ over $\mathbb{F}_2$

I just asked wolframalpha to factor $X^{16}+X$ over $\mathbb{F}_2$. The normal factorization is $$X(X+1)(X^2-X+1)(X^4-X^3+X^2-X+1)(X^8+X^7-X^5-X^4-X^3+X+1)$$ and over $GF(2)$ it is $$X(X+1)(X^2+X+1)(X^4+X+1)(X^4+X^3+1)(X^4+X^3+X^2+X+1).$$ Does the second form follow from the first, or is there a different way to factor over $\mathbb{F}_2$? I noticed that simply replacing the $-$ signs with $+$ signs in the first factorization doesn't yield the second one.

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The last term of the first is (mod $2$) the product of the $4$th and $5$th term of the second. The first $4$ terms of the first were irreducible mod $2$. –  André Nicolas Jul 12 '11 at 7:16

It shouldn't be a surprise that switching from integer (or rational) coefficients to modular coefficients allows for further factorization. The simplest example is probably the factorization $$x^2+1=x^2+2x+1=(x+1)^2$$ over $F_2$.

Here the key difference between the two factorizations is that the polynomial of degree 8 splits into a product of two quartic polynomials over $F_2$: $$(x^4+x+1)(x^4+x^3+1)=x^8+x^7+x^5+x^4+x^3+x+1$$ that is equal to (up to sign changes) your last factor.

Once you start on finite fields in your studies you will immediately learn that all the elements of $GF(16)$ are roots of the polynomial $x^{16}+x$. As that field is a degree 4 extension of $F_2$, all those elements have minimal polynomials of degree a factor of 4. Furthermore, all such irreducible polynomials appear as factors of $x^{16}+x$. Now, we easily see that both $x^4+x+1$ and its reciprocal polynomial $x^4+x^3+1$ are both irreducible in the ring $F_2[x]$, so they must appear as factors.

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The second form does follow from the first. Note that over $\mathbb F_2$, the pair of polynomials $$X^2 - X + 1 \quad \text{ and } \quad X^2 + X + 1$$ and the other pair $$X^4 - X^3 + X^2 - X + 1 \quad \text{ and } \quad X^4 + X^3 + X^2 + X + 1$$ are the same. The only difference that comes up in $\mathbb F_2$ is that the polynomial of degree $8$ factors.