Answer
$3x^3-7x^2+\frac{41}{2}x-\frac{123}{2}+\frac{{369}}{{2x+6}}
$
Work Step by Step
Re-arranging the order of the terms and leaving space for missing terms, the given is equivalent to
$$
(6x^4+4x^3-x^2+0x+0)÷(2x+6)
.$$
Using the long division method shown below, then
$$\begin{aligned}
&
(6x^4+4x^3-x^2)÷(2x+6)
\\&=
3x^3-7x^2+\frac{41}{2}x-\frac{123}{2}+\frac{\color{red}{369}}{\color{blue}{2x+6}}
.\end{aligned}
$$
