Answer
$z+9+\dfrac{39}{z-5}$
Work Step by Step
Setup: (the variable is $z$ 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.
$\begin{array}{rrr}
{5\ \ )} &{1}&{4}&{-6}\\
{ } &{ }&{ } &{ }\\
\hline &{1 }&{ } &{ }\end{array}$
Fill the next entries, column by column:
Middle row: k$\times$(previous bottom row entry)
Bottom row: add the two entries above.
Repeat.
$\begin{array}{rrr}
{5\ \ )} &{1}&{4}&{-6}\\
{ } &{ }&{5 } &{45 }\\
\hline &{1 }&{ 9} &{ 39}\end{array}$
Interpret result:
Q(x) is the quotient, R(x) the remainder. $\displaystyle \frac{P(x)}{D(x)}=Q(x)+\frac{R(x)}{D(x)}$
$Q(z)=z+9\quad R(p)=39$
$\displaystyle \frac{z^{2}+4z-6}{z-5}=z+9+\frac{39}{z-5}$