I have a quartic equation that is the determinant of a 4-by-4 matrix that looks like:
$det(M-\lambda I) = det \left( \matrix{m_{11}-\lambda & m_{12} & m_{13} & 0 \\ m_{21} & m_{22}-\lambda & m_{23} & 0 \\ 0 & m_{32}-1 & m_{33}-\lambda & m_{34} \\ 1 & 0 & 0 & -\lambda} \right)$
I know that some of the elements $m_{ij}$ are either positive or negative (e.g., $m_{21}$ is negative, $m_{22}$ is positive) and that I have 2 real distinct roots and 2 complex conjugate roots.
I am looking for a way to describe how the roots will vary as function of the various $m_{ij}$ given this restrictions.
Is there a way to decompose the quartic into a quadratic and two linear parts given what I know about my problem?
I have read Wikipedia's page and Wolfram's page but so far I have been unable to spot anything that could help me.
Maybe this simpler formulation might be of use. The general quartic,
$$x^4+ax^3+bx^2+cx+d=0$$
can be solved as,
$$x_{1,2} = -\tfrac{1}{4}a+\tfrac{1}{2}\sqrt{u}\pm\tfrac{1}{4}\sqrt{3a^2-8b-4u+\frac{-a^3+4ab-8c}{\sqrt{u}}}\tag1$$
$$x_{3,4} = -\tfrac{1}{4}a-\tfrac{1}{2}\sqrt{u}\pm\tfrac{1}{4}\sqrt{3a^2-8b-4u-\frac{-a^3+4ab-8c}{\sqrt{u}}}\tag2$$
where,
$$u = \frac{a^2}{4} +\frac{-2b+v_1^{1/3}+v_2^{1/3}}{3}$$
and the $v_i$ are the two roots of the quadratic,
$$v^2 + (-2 b^3 + 9 a b c - 27 c^2 - 27 a^2 d + 72 b d)v + (b^2 - 3 a c + 12 d)^3 = 0$$
Note: If you prefer only one cube root extraction, you can use the RHS of the relation,
$$v_1^{1/3}+v_2^{1/3} = \color{brown}{v_1^{1/3}}+\frac{b^2 - 3 a c + 12 d}{\color{brown}{v_1^{1/3}}}$$
Your quartic's two real roots will be given either by $(1)$ only, (or $(2)$ only) depending on which sign of $\pm\sqrt{u}$ you choose.