Quartic as a product of quadratics
After a linear transformation, one may assume that the quartic has the form $x^4 + a x^2 + b x + c$. As suggested, let's try to write
$$x^4 + a x^2 + b x + c = (x^2 + t)^2 - (r x + s)^2$$
over $\mathbf{R}$, which will give the desired factorization. In order to achieve this, it suffices to show that we can choose $t$ so that the difference
$$\Delta:=x^4 + a x^2 + b x + c - (x^2 + t)^2$$
has the following properties:
- The leading coefficient of $x^2$ is negative.
- There are repeated roots.
This implies it has the form $-(rx + s)^2$. The leading coefficient of $\Delta$ is $a - 2t$, so for condition 1, we need $2t > a$. For condition 2, the discriminant of $\Delta$ factors as
$$(2t-a)(8 t^3 - 4 a t^2 - 8 c t + 4 a c - b^2) =:(2t-a)g(t)$$
To summarize, it suffices to show that the second factor $g(t)$ above has a real root $2t > a$. But one finds:
$$g(a/2) = -b^2, \quad g(\infty) = + \infty.$$
So, as long as $b \ne 0$, we deduce that then $g(t)$ has a real root $2t > a$ by the intermediate value theorem, and we are done.
This still leaves the case when $b = 0$ which one can do by hand directly. In this case, we have $$x^4 + a x^2 + c = (x^2 + a/2)^2 - (a^2/4 - c),$$ which works if $a^2/4 \ge c$ and hence in particular when $c$ is negative, or $$x^4 + a x^2 + c = (x^2 + \sqrt{c})^2 - (2 \sqrt{c} - a) x^2,$$ which works in the complementary case $c \ge a^2/4$.