Answer
$p-4+\dfrac{9}{p+1}$
Work Step by Step
Setup: (the variable is $p$ 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.
$\begin{array}{rrr}
{-1)} &{1}&{-3}&{5}\\
{ } &{ }&{ } &{ }\\
\hline &{1 }&{ } &{ }\end{array}$
Fill the next entries, column by column:
Middle row: k$\times$(previous bottom row entry)
Bottom row: add the two entries above.
Repeat.
$\begin{array}{rrr}
{-1)} &{1}&{-3}&{5}\\
{ } &{ }&{-1 } &{ 4 }\\
\hline &{1 }&{ -4} &{ 9}\end{array}$
Interpret result:
Q(x) is the quotient, R(x) the remainder. $\displaystyle \frac{P(x)}{D(x)}=Q(x)+\frac{R(x)}{D(x)}$
$Q(p)=p-4,\quad R(p)=9$
$\displaystyle \frac{p^{2}-3p+5}{p+1}=p-4+\frac{9}{p+1}$
