Answer
$2y^{4}+y^{3}+3y^{2}+6y+12+\displaystyle \frac{13}{y-3}$
Work Step by Step
Setup:
Dividing with $y-3$,
place $3$ 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.
$\begin{array}{rrrrrrr}
{3\ \ )} &{2}&{-5}&{0}&{-3}&{6 }&{-23 }\\
{ } &{ }&{ }&{ }&{ } &{ }&{ }\\
\hline &{2 }&{ }&{ }&{ } &{ }&{ }\end{array}$
Fill the next entries, column by column:
Middle row: k$\times$(previous bottom row entry)
Bottom row: add the two entries above.
Repeat.
$\begin{array}{rrrrrrr}
{3\ \ )} &{2}&{-5}&{0}&{-3}&{-6 }&{-23 }\\
{ } &{ }&{6 }&{3 }&{9 } &{18 }&{36 }\\
\hline &{2 }&{1 }&{ 3}&{6 } &{12 }&{13 }\end{array}$
Interpret result:
Q(x) is the quotient, R(x) the remainder. $\displaystyle \frac{P(x)}{D(x)}=Q(x)+\frac{R(x)}{D(x)}$
$Q(y)=2y^{4}+y^{3}+3y^{2}+6y+12 \quad R(y)=13$
$\displaystyle \frac{2y^{5}-5y^{4}-3y^{2}-6y-23}{y-3}= $
$=2y^{4}+y^{3}+3y^{2}+6y+12+\displaystyle \frac{13}{y-3}$
