Answer
$x^{4}+2x^{3}+2x^{2}+7x+10+\displaystyle \frac{18}{x-2}$
Work Step by Step
Setup:
Dividing with $x-k$, place k as the top left entry.
List coefficients of the numerator (0 for missing powers) in the first row.
Copy the leading coefficient to the bottom row, same column.
$\begin{array}{rrrrrrr}
{2\ \ )} &{1}&{0}&{-2}&{3}&{-4 }&{-2 }\\
{ } &{ }&{ }&{ }&{ } &{ }&{ }\\
\hline &{1 }&{ }&{ }&{ } &{ }&{ }\end{array}$
Fill the next entries, column by column:
Middle row: k$\times$(previous bottom row entry)
Bottom row: add the two entries above.
Repeat.
$\begin{array}{rrrrrrr}
{2\ \ )} &{1}&{0}&{-2}&{3}&{-4 }&{-2 }\\
{ } &{ }&{2 }&{4 }&{4 } &{14 }&{20 }\\
\hline &{1 }&{2 }&{2 }&{7 } &{ 10}&{18 }\end{array}$
Interpret result:
Q(x) is the quotient, R(x) the remainder. $\displaystyle \frac{P(x)}{D(x)}=Q(x)+\frac{R(x)}{D(x)}$
$Q(x)=x^{4}+2x^{3}+2x^{2}+7x+10 \quad R(x)=18$
$\displaystyle \frac{x^{5}-2x^{3}+3x^{2}-4x-2}{x-2}= \\x^{4}+2x^{3}+2x^{2}+7x+10+\dfrac{18}{x-2}$
