Which way I can determine whether the polynomial $x^5 + 5x^2 +1$ is irreducible over $\mathbb Q$ or not?
Mod $p$ Irreducibility Test and Eisenstein's criterion not applicable here.
Which way I should proceed?
[Math] Polynomial $x^5 + 5x^2 +1$ is irreducible or not
abstract-algebrairreducible-polynomialspolynomialssubstitution
Related Solutions
With respect to your specific question (and quartics in general), the RRT doesn't solve the problem but it can help. Once you verify that your quartic has no roots, the only way it can factor is as a product of quadratic polynomials. Additionally, you can guarantee that the quadratic equations are monic and have integer coefficients. Observing that the constant terms multiply to $49$, we have four options:
$$x^4 + 2x^2 + 49 = (x^2 + ax + 7)(x^2 + bx + 7)$$
$$x^4 + 2x^2 + 49 = (x^2 + ax - 7)(x^2 + bx - 7)$$
$$x^4 + 2x^2 + 49 = (x^2 + ax + 1)(x^2 + bx + 49)$$
$$x^4 + 2x^2 + 49 = (x^2 + ax -1)(x^2 + bx - 49)$$
The vanishing of the cubic term gives $a = -b$ in each scenario, and the vanishing of the $x$ term eliminates scenarios 3 and 4. Comparing the $x^2$ terms in the first two scenarios gives:
$$14 - a^2 = 2$$
$$-14 - a^2 = 2$$
And it is easily seen that neither of these have solutions in the integers.
A couple of strategies:
This may be the only strategy you can use at this point (without Eisenstein's Criterion). If you are able to show that your polynomial (let's say $p(x)$) has no rational roots (by the rational root test for instance), then $p(x)$ has no linear factors in $\Bbb Q[x]$. Thus, since $p(x)$ is quartic, if it is to be reducible, then it must factor as two quadratics. This is because a quartic is either factored as: 4 linear, 1 linear and 1 cubic, 2 linear and 1 quadratic, or 2 quadratics, if it is reducible. So you suppose that $p(x)=(x^2+ax+b)(x^2+cx+d)$ for some $a,b,c,d\in\Bbb Z$. Then you compare the coefficients of the terms of $p(x)$ with the expanded form of that polynomial involving $a,b,c,d$ to see if there is a solution. If there cannot be $a,b,c,d$ satisfying these requirements, then $p(x)$ must be irreducible in $\Bbb Q[x]$.
You can use Eisenstein's Criterion here (although if you haven't learned about it this may be a problem). However, it will not work immediately. You can use the result (may require a proof) that $f(x)$ is irreducible if and only if $f(x+a)$ is irreducible for some $a\neq 0$ in the field you are working with. Thus, you can show that $f(x+a)$ is irreducible for some $a\in\Bbb Z$ by Eisenstein's Criterion, which will allow you to conclude that $f(x)$ is irreducible in $\Bbb Q[x]$.
Another way to show this polynomial is irreducible in $\Bbb Q[x]$, which is a more interesting approach (although not expected) is the following. Using the aforementioned theorem of $f(x)$ being irreducible if and only if $f(x+a)$ is irreducible, one can show that $f(x)=x^{p-1}+x^{p-2}...+x+1$ is irreducible in $\Bbb Q[x]$ for any positive prime $p$. One also needs to use Eisenstein's Criterion here. In this case, for $p=5$, we see that the polynomial in question is irreducible in $\Bbb Q[x]$.
Proof for strategy 3:
I will leave it to you to prove $f(x)$ is irreducible if and only if $f(x+a)$ is irreducible. Define $\phi_p(x)\in\Bbb F[x]$ as $\phi_p(x)=\frac{x^p-1}{x-1}$ for a positive prime $p$. Then we know that $\phi(x)=x^{p-1}+x^{p-2}+...+x+1$. Evaluating at $x+1$, we have: $$\begin{align}\phi_p(x+1)&=\frac{(x+1)^p-1}{x+1-1}\\&=\frac{1}{x}\sum_{n=0}^p\binom{p}{n}x^n-1\\&=x^{p-1}+\binom{p}{1}x^{p-2}+...+\binom{p}{p-2}x+\binom{p}{p-1}\end{align}$$ Note that $p\mid\binom{p}{k}$ for every $k\in\{1,2,...,p-1\}$, but $p\nmid 1$ and $p^2\nmid p=\binom{p}{p-1}$. Thus, by Eisenstein's Criterion and the fact that $f(x)$ irreducible iff $f(x+a)$ irreducible, we have that $x^{p-1}+x^{p-2}+...+x+1$ is irreducible in $\Bbb Q[x]$.
Best Answer
Try $x=y-1$. It should help by the Eisenstein's criterion.