$x^4 -10x^2 +1 $ is irreducible over $\mathbb Q$ I have seen the thread Show that $x^4-10x^2+1$ is irreducible over $\mathbb{Q}$ but this didn't really have a full solution.
Is it true that if it is reducible then it can be factored into a linear factor or quadratic factor in the form $x^2 - a$. Is $a$ in the rationals? And what exactly is a linear factor.
What I did was the following:
Let $y=x^2$. Then $y^2 - 10y +1=0 \iff y^2 -10y = -1$, add $25$ to both sides:
$$y^2 - 10y + 25 = -1 + 25 \iff  (y - 5)^2 = 24 \iff y = 5 \pm 2\sqrt6$$
So $x^2=5 \pm 2 \sqrt6= (\sqrt2 \pm \sqrt3 )^2$
So $x^4 -10x^2 +1 =(x^2 -(\sqrt2 + \sqrt3 )^2)(x^2 - (\sqrt2 - \sqrt3)^2)$
Then difference of two squares:
$ (x-(\sqrt2 + \sqrt3))(x+(\sqrt2 + \sqrt3))(x-(\sqrt2 - \sqrt3))(x+(\sqrt2 - \sqrt3))=0$
Expanding this out gives $x^4 -10x^2 +1 =0$ so we have found all the roots and we can confirm that they are the roots right? None of these roots are rational so it must be irreducible?
Please can you advise me on this method that I am using and not direct me to a theorem or any type of shortcut. 
Thanks in advance.
 A: Adding a method using the roots that you found directly (I deleted my other answer, and relocated it to the companion thread because it properly belongs there, but is misplaced here).
You have correctly identified that the zeros of
$$
p(x)=x^4-10x^2+1
$$
are $x_1=\sqrt2+\sqrt3$, $x_2=\sqrt2-\sqrt3$, $x_3=-\sqrt2+\sqrt3$ and $x_4=-\sqrt2-\sqrt3$. Therefore over the reals we have the factorization
$$
p(x)=(x-x_1)(x-x_2)(x-x_3)(x-x_4).
$$
Others have already explained to you why it is not sufficient to check that none of the roots are rational - namely that the polynomial could still have quadratic factors with rational coefficients. But, we can take advantage of the list of zeros to exclude that possibility as well.
If $p(x)=f(x)g(x)$ were a factorization as a product of two quadratics with rational coefficients, then $x_1$ must be a zero of one of the factors. Without loss of generality we can assume that $f(x_1)=0$. This means that the other zero of $f(x)$ must be either $x_2,x_3$ or $x_4$. But we can check that none of
$$
\begin{aligned}
(x-x_1)(x-x_2)&=(x-\sqrt2)^2-(\sqrt3)^2=x^2-2\sqrt2x-1\\
(x-x_1)(x-x_3)&=(x-\sqrt3)^2-(\sqrt2)^2=x^2-2\sqrt3x+1\\
(x-x_1)(x-x_4)&=x^2-(\sqrt2+\sqrt3)^2=x^2-5-2\sqrt6
\end{aligned}
$$
have rational coefficients. Therefore $p(x)$ has no quadratic factors with rational coefficients, and hence must be irreducible.
A: $(x^2+1)(x^2+2)$ has no rational roots, but is clearly reducible over $\mathbb{Q}[x]$, so your method is incorrect. In general the factors of a reducible polynomial may not be linear, so it may be reducible without having roots in the base field.
A: It is not true that if a polynomial of the form $x^4 + b x^2 + c \in \Bbb Q[x]$ is a product of two quadratics in $\Bbb Q[x]$ then the quadratics have the form $x^2 + a$. Consider, for example,
$$x^4 + x^2 + 1 = (x^2 + x + 1)(x^2 - x + 1).$$ (These quadratics both have negative discriminant and so are irreducible, so this is the only factorization into [monic] quadratics.) It is true, however, that since any polynomial of the above form is even, if it is a product of two quadratics, there is a factorization of the form
$$(x^2 + a x + b)(x^2 - a x + b).$$ In the case of $$f(x) := x^4 - 10x^2 + 1$$ expanding and solving for $a, b$ quickly leads to a contradiction.
Note that by the Rational Root Theorem, any rational root of $f(x)$ is either $\pm 1$, but neither of these are roots, so $f(x)$ has no linear factors, and hence it is irreducible.
Note, by the way, that the irrationality of all of the roots of a polynomial does not imply that the polynomial is irreducible, at least for polynomials of degree $\geq 4$. Consider the simple case $(x^2 - 2)^2$, which is visibly reducible but only has (irrational) roots $\pm \sqrt{2}$.
A: You have correctly identified the roots.
You may want to show that none of them are rational, so you know your polynomial has no rational factor of degree 1.
It could still be the product of two rational polynomials of degree $2$. To rule this out, prove that the three products of a fixed linear factor, $x-(\sqrt{2} + \sqrt{3})$, say, by each of the other three linear factor never gives a rational polynomial.

To answer your other questions, a linear factor is a factor of degree one. And a generic quadratic factor is of the form $x^{2} + a x + b$. For instance the polynomial
$$
(x^{2} + x + 1)^{2}
$$
can only be written as the product of irreducible, monic, rational polynomials as
$$
(x^{2} + x + 1) \cdot
(x^{2} + x + 1),
$$
and none of them has the form $x^{2} - a$.
A: First note that it suffices to show the polynomial is irreducible over $\mathbb{Z}$. We then note that $p(x)=x^4-10x^2-1$ splits either as
$$(x-a)(x^3+bx^2+cx+d)$$
or as
$$(x^2+bx+c)(x^2+dx+e)$$
In the first case $a$ would be a root of $p$, but by the rational root theorem the polynomial has no roots in $\mathbb{Q}$ (and hence not in $\mathbb{Z}$). In the second case
$$(x^2+bx+c)(x^2+dx+e)=x^4+(b+d)x^3+(e+c+bd)x^2+(be+dc)x+ec$$
so $e=c=\pm 1$, $b=-d$, filling this in we find
$$x^4+(b+d)x^3+(e+c+bd)x^2+(be+dc)x+ec=x^4+(2-b^2)+1$$
and $2-b^2=-10$ has no solution in $\mathbb{Z}$.
A: It's quite easy to find a factorization over $\mathbb{R}$:
\begin{align}
p(x)=x^4-10x^2+1
&=(x^2-1)^2-8x^2 \\[6px]
&=(x^2-2\sqrt{2}\,x-1)(x^2+2\sqrt{2}\,x-1) \\[6px]
&=(x-(\sqrt{2}+\sqrt{3}))(x-(\sqrt{2}-\sqrt{3}))\\
&\qquad \cdot (x-(-\sqrt{2}+\sqrt{3}))(x-(-\sqrt{2}-\sqrt{3}))\\
\quad
\end{align}
Our task is to show that $p(x)$ is the minimal polynomial of $\sqrt{2}+\sqrt{3}$, which implies it is irreducible. If not, the degree of $\sqrt{2}+\sqrt{3}$ over $\mathbb{Q}$ would be less than $4$. However
$\mathbb{Q}(\sqrt{2}+\sqrt{3})=\mathbb{Q}(\sqrt{2},\sqrt{3})$ (prove it), which has degree $4$ over $\mathbb{Q}$, because $\sqrt{3}\notin\mathbb{Q}(\sqrt{2})$.
