In our homework, we are supposed to find all the irreducible polynomials of $\mathbb F_2[X]$ with degree $\leq 4$. Ok, I know that there are at least two helpful threads dealing with this question: Find all irreducible monic polynomials in $\mathbb{Z}/(2)[x]$ with degree equal or less than 5
Find all irreducible polynomials of degrees 1,2 and 4 over $\mathbb{F_2}$.
(1) According to the first link: "A polynomial of degree $2$ or $3$ is irreducible if and only if does not have linear factors."
Why? Let's consider $X^2+X=X(X+1)$. $X^2+X$ is not a unit, and neither are $X$ or $X+1$, as I see it. So why then isn't $X^2+X$ irreducible? [Let $R$ be a commutative ring. Then $r\in R$ is called irreducible if $r\ne 0, r\notin R^{\star}$ and for every decomposition $r =ab\Rightarrow a\in R^{\star} \vee b\in R^{\star}$.]
Kind regards, MathIsFun
Your reasoning already shows that $X^2+X$ is not irreducible: Irreducibility means that there are no factorizations $r=ab$ with both $a$ and $b$ a non-unit. As you already stated, neither $X$ nor $X+1$ are units, so their product is reducible.
In general: Given a polynomial $f\in\mathbb{F}_2[X]$ of degree $2$ or $3$. Any decomposition into factors $f=ab$ has to satisfy $\deg(f)=\deg(a)+\deg(b)$. Suppose $\deg(a)\leq\deg(b)$, then we have two cases:
1)$\deg(a)=0$, but then $a\in\mathbb{F}_2^\ast$ and hence $a$ is a unit
2)$\deg(a)=1, \deg(b)=1,2$: Both factors are non-units, hence $f$ is reducible
So, $f$ is irreducible if and only if it does not factor a linear factor. Note that this reasoning works only for polynomial rings over fields and for polynomials of degree $\leq 3$.