Is there any systematic way to factorise expressions of the form $ n^4+pn^2 +q$ Is there any systematic way to factorise expressions of the form $ n^4+pn^2 +q$

Expression like $n^4+n^2+1$ usually become helpful for telescopic type , when used in the form$(n^2+n+1)(n^2-n+1)$, but for an expression like $n^4+pn^2 +q= (n^2+an+b)(n^2+cn+d)$ systematic way to factorise it like above (if possible).
Yes, of course! Always there exist a way to factorize it.

If $p^2-4q\geq0,$ so we can use a way, which is a similar to the following:
$$x^4+4x^2+3=x^4+3x^2+x^2+3=(x^2+1)(x^2+3).$$
But for $p^2-4q&lt;0$ there is a way, which is a similar to the following:
$$ x^4+x^2+25=x^4+10x^2+25-9x^2=(x^2-3x+5)(x^2+3x+5) $$

For $ p^2-4q&lt;0 $ in the general case we obtain:
$$ x^4+px^2+q=(x^2+\sqrt{q})^2-2x^2\sqrt{q}+px^2=(x^2+\sqrt{q})^2-(2\sqrt{q}-p)x^2=$$
$$=\left(x^2-\sqrt{2\sqrt{q}-p}x+\sqrt{q}\right)\left(x^2+\sqrt{2\sqrt{q}-p}x+\sqrt{q}\right) $$
This is a quadratic in $n^2,$ so you can always factorize it over the complex numbers.