A polynomial $p(x)$ is irreducible over a field $K$ iff $kp(x)$ is also irreducible over $K$, where $k \in K$.
Proof. We will prove both directions.
(i) $p(x)$ is irreducible over $K \Rightarrow kp(x)$ is irreducible over $K$.
Suppose $kp(x)$ is reducible in $K$, i.e., $kp(x) = f(x)\cdot g(x)$ for some $f,g \in \mathcal{P}^n_{K}$ [the set of all polynomials of degree $< n$ over $K$]. This implies that $p(x) = (k^{-1}f(x))\cdot g(x)$, since $k^{-1} \in K$.
This means that $p(x)$ is reducible, which is a contradiction. Hence $kp(x)$ is irreducible over $K$.
(ii) $kp(x)$ is irreducible over $K \Rightarrow p(x)$ is irreducible over $K$.
Suppose $p(x)$ is reducible in $K$, i.e., $p(x) = f(x)\cdot g(x)$ for some $f,g \in \mathcal{P}^n_{K}$. This implies that $kp(x) = (kf(x))\cdot g(x)$, since $k \in K$.
This means that $kp(x)$ is reducible, which is a contradiction. Hence $p(x)$ is irreducible over $K$.
Note. The set of all polynomials of degree $<n$ over a Galois field $GF(p)$ is denoted as $GF(p^n)$, and is called an extended Galois field.
Just like integers, polynomials over a ring can also be divided, and Euclid’s Division Lemma applies to them as well. i.e., given any two polynomials $g(x), r(x) \in K[X]$, there exist unique $q(x), r(x) \in K[X]$ such that
and $0 \leq$ deg $r <$ deg $g$. Given this, we can apply Euclid’s Algorithm and Euclid’s Extended Algorithm to polynomials over $K$, and thus find their greatest common divisors and inverses.
Euclid’s Algorithm for polynomials is exactly analogous to Euclid’s Algorithm for integers:
gcd$(a(x),b(x))$
Again, Euclid’s Extended Algorithm for polynomials is exactly analogous to Euclid’s Extended Algorithm for integers:
inverse$(a(x),b(x))$