Irreducible Polynomial in $\mathbb{Q}[x]$

Question

Show that $x^4+2x^2+4$ is irreducible in $\mathbb{Q}[x]$.

Answer 1

If it had a factor that is a linear polynomial in $\mathbb{Q}$ then it would have a rational root. The only possibilities for rational roots are divisors of $4$: $\pm1$, $\pm2$, or $\pm4$. We can check if these are solutions.

If it factors as $(x^2+ax+b)(x^2+px+q)=x^4+(a+p)x^3+(b+q+ap)x^2+(bp+aq)x+bq$

So, we get $$\begin{align}a+p&=0\\b+q+ap&=2\\pb+aq&=0\\bq&=4\end{align}$$

If $a=0$, then $p=0$, and the solutions for $b,q$ are not rational.

If $a\neq0$ then $p=-a$, $b=q=\pm2$, and there are no rational solutions for $a,p$.

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Alternatively, we are lucky, we can find all the roots of this polynomial. Notice that if we put $y:=x^2$ the polynomial reads $$y^2+2y+4$$. We can find the roots of this polynomial and with them find the values for $x$. With these values we can write $$x^4+2x^2+4=(x-r_1)(x-r_2)(x-r_3)(x-r_4)$$ and then see if any factor $(x-r_i)$ or $(x-r_i)(x-r_j)$ is a rational polynomial.