Irreducibility of a polynomial

Rational root test. Possible rational roots: $\pm1$.

Test. $P(1)=1+1+1=3\neq0$; $P(-1)=-1-1+1=-1\neq0$.

Conclusion. No rational root, so $P$ has no linear factor over $\mathbb{Q}$. Since $P$ has degree $3$, if it were reducible it would have a factor of degree $1$ or $2$, hence a linear factor. So $P$ is irreducible over $\mathbb{Q}$.

Remark. The polynomial has one real root (by the intermediate value theorem, since $P(-1)<0$ and $P(0)>0$) and two complex conjugate roots.