Answer
$4a^{2}+a+3$
Work Step by Step
Setup: (the variable is $a$ instead of $x$).
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.
$\begin{array}{rrr}
{1\ \ )} &{4}&{-3}&{2}&{-3}&{ }\\
{ } &{ }&{ }&{ }&{ } &{ }\\
\hline &{4 }&{ }&{ }&{ } &{ }\end{array}$
Fill the next entries, column by column:
Middle row: k$\times$(previous bottom row entry)
Bottom row: add the two entries above.
Repeat.
$\begin{array}{rrr}
{1\ \ )} &{4}&{-3}&{2}&{-3}&{ }\\
{ } &{ }&{4 }&{ 1}&{ 3} &{ }\\
\hline &{4 }&{1 }&{ 3}&{0 } &{ }\end{array}$
Interpret result:
Q(x) is the quotient, R(x) the remainder. $\displaystyle \frac{P(x)}{D(x)}=Q(x)+\frac{R(x)}{D(x)}$
$Q(a)=4a^{2}+a+3\quad R(a)=0$
$\displaystyle \frac{4a^{3}-3a^{2}+2a-3}{a-1}= 4a^{2}+a+3$
