Is $f(X)=X^4-22X^2+1\in\mathbb{Q}[X]$ irreducible?
The solution proved that it is irreducible. However, it says that if $f$ factors in $\mathbb Z[X]$, then either it can be factored to a quadratic or has a linear factor. And that the only possible linear factors are $X + 1$ and $X - 1$. I can't see how these linear factors were found.
Any idea is appreciated.
On
In addition to Arthur's answer, we can see that $X^4-22X^2+1$ cannot be factored into two quadratics over $\mathbb{Q}$. Over $\mathbb{F}_{11}$, $$X^4-22X^2+1=(X^2-1)^2-(3X)^2=(X^2-3X-1)(X^2+3X-1)\,.$$ Observe that $X^2-3X-1$ and $X^2+3X-1$ are irreducible over $\mathbb{F}_{11}$, as $$\left(\frac{3^2+4}{11}\right)=\left(\frac{2}{11}\right)=\left(\frac{-9}{11}\right)=\left(\frac{-1}{11}\right)\,\left(\frac{3}{11}\right)^2=-1\,.$$ Therefore, if $X^4-22X^2+1$ can be factored over $\mathbb{Q}$ into $X^2+aX+b$ and $X^2+cX+d$ for some $a,b,c,d\in\mathbb{Z}$, we must have $b=d=-1$, and $a=-c\equiv 3\pmod{11}$. This shows $a$ is an integer with the property that $$-22=-2-a^2\,,\text{ whence }a^2=20\,,$$ which is absurd. (We do not actually need to look at $X^4-22X^2+1$ modulo $11$, but it helps eliminate the case where $b=d=+1$.)
The reason we are able to restrict it to $x+1$ and $x-1$ is the rational root theorem which says, among other things, that if a monic integer polynomial has rational roots, then its roots are integers and they divide the constant term. Since the constant term is $1$, the only possible rational roots the polynomial could have are $\pm 1$, which gives those possible linear factors.