How do I continue solving this polynomial division? I had this equation as a bonus question on a quiz today. I got to a certain extend and was completely unsure how to continue.


*$$(10x^4-18x^3-94x^2-8x+2) \div (10x+2)$$


I figured the easiest first step would be to divide. (Actually, I assumed it was factorable, but I have no idea how to go about that)
$$\frac{(10x^4-18x^3-94x^2-8x+2)} {(10x+2)}$$
$$\frac{10x^4}{10x+2} + \frac{-18x^3}{10x+2} + \frac{-94x^2}{10x+2} + \frac{-8x}{10x+2} + \frac{2}{10x+2}$$
Simplifying some more (multiplying by $10x+2$)
$$10x^4-18x^3-94x^2-8x-2 = 0$$
But what do I do from here? Do I factor? Do do something else?
If I do factor How do I go about that? If not, what do I do instead?

Polynomial long division:
\begin{array}{c|cc}
 & &\color{red}{x^3} & \color{green}{-2x^2}& \color{blue}{-9x}& \color{orange}{+1}& \\  \hline
10x+2 & 10x^4 & -18x^3  & -94x^2& -8x& +2  \\
& \color{red}{10x^4} &\color{red}{+2x^3}\\
& & -20x^3  & -94x^2& \vdots & \vdots\\
& & \color{green}{-20x^3}  & \color{green}{-4x^2}\\
& &  & -90x^2& -8x & \vdots\\
& &  & \color{blue}{-90x^2}& \color{blue}{-18x} \\
& &  & & 10x & +2\\\
& &  & & \color{orange}{10x} & \color{orange}{+2}\
\end{array}
$\therefore (10x^4-18x^3-94x^2-8x+2) =(10x+2)(x^3-2x^2-9x+1)$
