The book I am reading provides the following example to show how irreducibility works:
Let $f(x) = \frac{3}{7}x^4- \frac{2}{7}x^2+\frac{9}{35}x+\frac{3}{5}$. We will show that $f(x)$ is irreducible over $\mathbb{Q}$. First, let $h(x) = 35f(x) = 15x^4-10x^2+9x+21$. Then $f(x)$ is irreducible over $\mathbb{Q}$ if $h(x)$ is irreducible over $\mathbb{Z}$. Next, applying the Mod 2 Irreducibility Test to $h(x)$, we get $\bar{h}(x)=x^4 + x + 1$. Clearly, $\bar{h}(x)$ has no zeros in $\mathbb{Z}_2$. Furthermore, $\bar{h}(x)$ has no quadratic factor in $\mathbb{Z}_2[x]$ either. [~an argument here about why this is true~] Thus, $\bar{h}(x)$ is irreducible over $\mathbb{Z}_2[x]$. This guarantees that $h(x)$ is irreducible over $\mathbb{Q}$.
My question: Why is it obvious that I don't need to check degree $3$ polynomials on $\bar{h}$?
They skip mentioning this, and the only thing I could think was to divide $\bar{h}$ by all $8$ degree $3$ polynomials and check.
Edit: I think if I check degree $1$'s, I get degree $3$'s for free. But I don't know how to show this to myself.
We do not need to check for degree three factors, since then the other factor would be linear, i.e., we would have a root, which we obviously do not have: $0^4+0+1=1$ and $1^4+1+1=1$. Indeed, having a cubic factor means that $x^4+x+1=(x^3+ax^2+bx+c)(x+d)$ for some $a,b,c,d\in \mathbb{F}_2$.