TITLE: Why is $X^4-16X^2+4$ irreducible in $\mathbb{Q}[X]$?
QUESTION [5 upvotes]: Determine whether $X^4-16X^2+4$ is irreducible in $\mathbb{Q}[X]$.
To solve this problem, I reasoned that since $X^4-16X^2+4$ has no rational roots hence irreducible.
But there is a hint to this question that uses different approach:
Try supposing it is reducible, then it must factor into a product of two monic quadratic polynomials with integer coefficients. Then show that it is impossible, then conclude the original polynomial is irreducible.
My questions regarding the hints:
1. Why can't we just show that since $X^4-16X^2+4$ has no rational roots hence irreducible?
2. Why do we have to factorise into a product of two monic quadratic polynomials with integer coefficients? Why monic? And why can't we factorise into polynomial with degree 1 and 3? Also, lastly why the coefficients have to be integers?
Thank you for any explanations!

Edit: Thanks for all the answers.
I saw that Second Gauss Lemma is used. But I only learned the first one in class. Which is:
Let $R$ be a Unique Facorisation Domain. If $f,g\in\mathbb{R}[X]$ are primitive, then so too is their product $fg$.
Is it inevitable to use Second Gauss Lemma? Is there any other way around?

REPLY [5 votes]: By the rational root theorem, no root is rational. We only need check a factorisation $x^4-16x^2+4=(x^2+ax+b)(x^2+cx+d)$ with rational coefficients doesn't exist. Solving $a+c=0,\,ad+bc=0,\,bd=4$ (from the coefficients for $x^3,\,x,\,1$) gives $c=-a,\,d=-b=\pm 2i$, a contradiction.

REPLY [2 votes]: The roots of your polynomial are $\pm\sqrt{8\pm2\sqrt{15}}$. Since none of them is rational, the only way that your polynomial could be reducible in $\mathbb Q[x]$ was if you could take two roots $\alpha$ and $\beta$ among those four roots and then get $(x-\alpha)(x-\beta)\in\mathbb{Q}[x]$. You can easily check that that will not take place.