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**ThePerfectHacker** First of all $\displaystyle \alpha \not = 0$! Now define $\displaystyle \phi: \mathbb{Q}[X]\to \mathbb{Q}[X]$ by $\displaystyle \phi (f(X)) = f(\alpha X+\beta)$. Since $\displaystyle \alpha\not = 0$ it means $\displaystyle \deg f = \deg \phi (f)$. Say that $\displaystyle f(X) = f_1(X)f_2(X)$ where $\displaystyle \deg f_1,\deg f_2 < \deg f$ and $\displaystyle f_1,f_2\not \in \mathbb{Q}$. This means, $\displaystyle \phi (f) = \phi (f_1)\phi (f_2)$ but $\displaystyle \deg \phi (f) = \deg (f)$, $\displaystyle \deg \phi (f_1) = \deg f_1$, $\displaystyle \deg \phi (f_2) = \deg f_2$. It follows that $\displaystyle \deg \phi(f_1),\deg \phi(f_2) < \deg \phi (f)$ and $\displaystyle \phi(f_1),\phi(f_2) \not \in \mathbb{Q}$, therefore $\displaystyle \phi (f) = f(\alpha X + \beta)$ is not irreducible. We have proved that if $\displaystyle f(X)$ is reducible then $\displaystyle f(\alpha X + \beta)$ is reducible, taking the contrapositive we have proven that if $\displaystyle f(\alpha X + \beta)$ is irreducible then $\displaystyle f(X)$ is irreducible. To prove the converse I am going to give you a hint (Nod). Define $\displaystyle \theta : \mathbb{Q}[X] \to \mathbb{Q}[X]$ by $\displaystyle \theta (g(X)) = g\left( \tfrac{1}{\alpha}X - \tfrac{\beta}{\alpha}\right)$ (again $\displaystyle \alpha\not = 0$). Notice that $\displaystyle \phi (\theta(g(X))) = \theta (\phi (g(X))) = g(X)$.