Proving that $x^{2}+2$, $x^{2}-x+4$, and $x^{3}+3x-1$ are irreducible over $\mathbb{Q}$ 
Let $f$, $g$ and $h$ be the polynomials given by:
  $$f(x)=x^{2}+2$$
  $$g(x)=x^{2}-x+4$$
  $$h(x)=x^{3}+3x-1$$
  Show that $f$, $g$ and $h$ are irreducible over $\mathbb{Q}$.

I do this:
$$x^{2}+2=(\sqrt{2}-i x) (\sqrt{2}+i x)$$
then it is irreducible over $\mathbb{Q}$.
$$x^{2}-x+4=4+(-1+x)(x)$$
then it is irreducible over $\mathbb{Q}$.
$$x^{3}+3x-1=-1+x (3+x^2)$$
then it is irreducible over $\mathbb{Q}$.
Is this true?
Thanks very much.
 A: We shall employ this criterion by Eisenstein.
This criterion applies immediately to $f(x)$ with $p=2$. Also, in writing $g(x+3)=x^2+5x+10$, we could apply the criterion to $g(x)$ with $p=5$. Finally, writing $h(x+1)=x^3+3x^2+6x+3$, we again apply the criterion to $h(x)$ with $p=3$. This finishes the proof.
P.S. we also notice here that a polynomial $f(x)$ is reducible if and only if $f(x+c)$ is so for every constant $c$. Hence the above reasonings.  
A: You can also try the following alternative method: if $\,p(x)\in\Bbb Q[x]\,$ is reducible, then it must be reducible over any finite prime field field $\,\Bbb F_p\cong \Bbb Z/p\Bbb Z\,$ . So take
$$(1)\;\;x^2+2\pmod 5 \;\;\text{(hint:}\,\,\Delta:=b^2-4ac=-8=2\pmod 5\,\text{... not a square in }\,\,\Bbb F_5 )$$
$$(2)\;\;\;\;\;\;\;\;x^2-x+4=x^2+6x+4\pmod 7\,\,(\Delta=20=6=-1\pmod 7)$$
$$(3)\;\;\;\;\;\;\;\;\;\;\;\;\;\;x^3+3x-1=x^3+x+1\pmod 2\ldots$$
A: Any non-trivial factorization of those polynomials would contain a linear factor, implying a rational root. The rational root test easily shows there are no rational roots. 
