This is a long division problem (you'll also find a,b and c as a bonus). In this case r will be your final remainder
Since it's a nightmare to do on latex I'll link to Wolfram: http://www.wolframalpha.com/input/?i...29%2F%28x-4%29
$\displaystyle \displaystyle \frac{3x^3-2x^2+2x-2}{x-4} = ax^2+bx+c+\frac{r}{x-4}$
$\displaystyle \displaystyle \Rightarrow 3x^3-2x^2+2x-2 = (x-4)(ax^2+bx+c)+r$
$\displaystyle \displaystyle \Rightarrow 3x^3-2x^2+2x-2 = ax^3+(b-4a)x^2+(c-4b)x-4c+r$
Comparing the coefficients:
It's obvious that $\displaystyle a = 3$.
Therefore $\displaystyle b-4a = -2 \Rightarrow b = 10$.
So $\displaystyle c-4b = 2 \Rightarrow c = 42$.
Finally $\displaystyle -4c+r = -2 \Rightarrow r = 166.$