Answer
$-4r^{5}-7r^{4}-10r^{3}-5r^{2}-11r-8-\displaystyle \frac{5}{r-1}$
Work Step by Step
Setup:
Dividing with $r-1$,
place $1$ 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.
$\begin{array}{rrrrrrrr}
{1\ \ )} &{-4}&{-3}&{-3}&{5}&{-6 }&{3 }&{3 }\\
{ } &{ }&{ }&{ }&{ } &{ }&{ }&{ }\\
\hline &{-4 }&{ }&{ }&{ } &{ }&{ }&{ }\end{array}$
Fill the next entries, column by column:
Middle row: k$\times$(previous bottom row entry)
Bottom row: add the two entries above.
Repeat.
$\begin{array}{rrrrrrrr}
{1\ \ )} &{-4}&{-3} &{-3} &{5} &{-6 } &{3 }&{3 }\\
{ } &{ } &{-4 } &{-7 } &{ -10 }&{-5 }&{-11 }&{-8 }\\
\hline &{-4 }&{ -7 }&{-10 }&{ -5} &{ -11}&{-8 }&{-5 } \end{array}$
Interpret result:
Q(x) is the quotient, R(x) the remainder. $\displaystyle \frac{P(x)}{D(x)}=Q(x)+\frac{R(x)}{D(x)}$
$Q(r)=-4r^{5}-7r^{4}-10r^{3}-5r^{2}-11r-8 \quad R(r)=-5$
$\displaystyle \frac{-4r^{6}-3r^{5}-3r^{4}+5r^{3}-6r^{2}+3r+3}{r-1}=$
$=-4r^{5}-7r^{4}-10r^{3}-5r^{2}-11r-8-\displaystyle \frac{5}{r-1}$
