Let $F$ be a field and $a,b\in F$ with $a\ne0$. Then, $f(x)\in F[x]$ is irreducible if and only if $f(ax+b)\in F[x]$ is irreducible.
This is my proof
$(\Rightarrow)$ Suppose $f(x)=h(x)g(x)$ is irreducible with $h(x),g(x)\in K[x]$, then $f(ax+b)=h(ax+b)g(ax+b)$. Because $f(x)$ is irreducible, then $h(x)\ne0$ or $g(x)\ne0$, then $h(ax+b)\ne0$ or $g(ax+b)\ne0$. So $f(ax+b)$ is irreducible. $(\Leftarrow)$ Suppose $f(ax+b)$ is irreducible, let $y=ax+b\in F$, then for every $c,d\in F$ with $c\ne0$, $f(cy+d)$ is irreducible. Put $c=a^{-1}$ and $b=-(a^{-1}b)$, we have $f(cy+d)=f(a^{-1}ax+a^{-1}b-a^{-1}b)=f(x)$.
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