A Family of Cubics

$\begin{aligned} f(x) & = ax^3+bx^2+cx+d \\ \implies f(1) & = a+b+c+d \\ 2k & = a+b+c+d \end{aligned}$

If $f(x)$ has the factor $ax^2+(c-d)x+d$ , we can assume:

$\begin{aligned} f(x) & = (x-g)(ax^2+(c-d)x+d) \\ & = ax^3 + (c-d-ag)x^2 + (d-cg+dg)x -dg \end{aligned}$

Equating coefficients, from the constant term, we have $\color{#3D99F6}{g=-1}$ ; and from the $x^2$ term,

$\begin{aligned} b & = c-d-a\color{#3D99F6}{g} & \small \color{#3D99F6}{g=-1} \\ b+d & = a+c \\ a+b+c+d & = 2a+2c \\ 2k & = 2(a+c) \end{aligned}$

$\ \implies \boxed{\text{Yes, }k=a+c}$

this facebook video explains the Game of G-filtered Polycules for Cubics; leave a comment.