Answer
$6x^{4}+12x^{3}+22x^{2}+48x+93+ \displaystyle \frac{187}{x-2}$
Work Step by Step
Dividing with $x-c\qquad...\qquad c=2.$
Don't forget to place 0's for the missing terms.
$\begin{array}{rrrrrrrr}
\underline{2}| &6 & 0 &-2 & 4 &-3 &1 \\
& & 12 &24 &44 &96 &186 \\
& --& --&--&-- &--&--\\
&6 & 12 &22 &48 &93 & 187 \end{array}$
Quotient = $6x^{4}+12x^{3}+22x^{2}+48x+93$
Remainder = $187$
$\displaystyle \frac{dividend}{divisor}=quotient+\frac{remainder}{divisor}$
$\displaystyle \frac{6x^{5}-2x^{3}+4x^{2}-3x+1}{x-2}$ = $6x^{4}+12x^{3}+22x^{2}+48x+93+ \displaystyle \frac{187}{x-2}$
