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