Is the polynomial $x^4+10x^2+1$ reducible over $\mathbb{Z}[x]$? Is the polynomial  $x^4+10x^2+1$  reducible over  $\mathbb{Z}[x]$?
 A: We see that our polynomial has no integer roots and has no real roots. So if it is reducible, there are integers $a$ and $b$ for which
$$x^4+10x^2+1=(x^2+ax+1)(x^2+bx+1).$$ 
Can you end it now? 
A: It doesn't have any rational roots, by the rational roots test.
To see whether $f(x) = x^4+10x^2 + 1 = (x^2+ax+b)(x^2+cx+d)$ or not, try to solve the system of equations $a+c = 0$, $ac+b+d = 0$, $bd = 1$ over the integers.
Alternately, you could find the roots of $f(x)$ (they are square roots of the roots of $y^2+10y+1$) and check whether any of them has degree 2 over $\mathbb Q$.
A: Hey whenever the expression contains even powers of $x$ you can always take $x^2=t$, to reduce the equation to :
$$t^2+10t+1$$ which is a quadratic polynomial over $t$
Now follow the usual factorization as you would for the quadratic expression....
$$t^2+10t+1=(t- \alpha)(t- \beta)$$ where
$$\alpha =\frac{-10+\sqrt{96}}{2}$$
$$\beta=\frac{-10-\sqrt{96}}{2}$$
Now put back $t=x^2$ and now you will see that the factors will lead to two more factors
$$(x^2-\alpha)(x^2-\beta)$$
which will give you 4 factors:
$$(x-\alpha)(x+\alpha)(x-\beta)(x+\beta)$$
Hope this helps …...
