Polynomial is irreducible over $\mathbb{Q}$
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 $\Bbb Q[x]$.
I suppose you have shown that the polynomial has no rational root. Now if the polynomial is reducible, then its factors must be two irreducible quadratic polynomials. Recognise that the polynomial is $\frac{x^5-1}{x-1}$. The roots of $x^4+x^3+x^2+x+1$ would be the roots of unity other than 1. Find the pair of irreducible quadratic polynomials and see that this factorisation is not possible in rational numbers.
There is no general strategy for testing whether a polynomial is irreducible except when the field is algebraically closed. It is also more difficult to check whether a polynomial is irreducible in $\Bbb Q$ than in $\Bbb R$.
You may check through linear algebra that your polynomial is the minimal polynomial of $\alpha=e^{\frac{2\pi i}{5}}$ over $\mathbb{Q}$. $\alpha^5-1=0$ and $\alpha-1\neq 0$, hence $\alpha$ is a root of
$$ \Phi_5(x)=\frac{x^5-1}{x-1} = x^4+x^3+x^2+x+1.$$
$\mathbb{Q}[x]/(\Phi_5(x))$ is a vector space over $\mathbb{Q}$ with dimension $4$ and a base given by $1,x,x^2,x^3$.
Since the following matrix
$$ \begin{pmatrix}0 & 0 & 0 & -1 \\ 1 & 0 & 0 & -1 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & -1\end{pmatrix} $$
representing the multiplication by $x$ in the previous ring, has full rank, $\Phi_5(x)$ is the minimal polynomial of $\alpha$ over $\mathbb{Q}$ and $\Phi_5(x)$ is an irreducible polynomial.