Answer
$x^{2}+5x+3$
Work Step by Step
When dividing a polynomial $f(x)$ with $(x-k)$
We set up syntehetic division $\quad \text{divisor } )\overline{\text{ dividend }}$,
by placing $k$ in place of the divisor,
and listing ALL coefficients of $f(x)$ (including the zeros ), starting from highest power of x.
$k=-2$
$ \begin{array}{lllllll}
& & -- & -- & -- & -- & \\
-2 & ) & 1 & 7 & 13 & 6 & \\
& & & & & & \\
& & -- & -- & -- & -- & \\
& & & & & & \\
& & & & & &
\end{array}$
We are ready.
Bring down the leading coefficient, $1$
Multiply $(-2)(1)=-2$ and place it in the next free slot of the middle row.
$ \begin{array}{lllllll}
& & -- & -- & -- & -- & \\
-2 & ) & 1 & 7 & 13 & 6 & \\
& & & -2 & & & \\
& & -- & -- & -- & -- & \\
& & 1 & & & & \\
& & & & & &
\end{array}$
Add $7+(-2)=5$ and place $5$ in the bottom row.
Multiply $(-2)(5)=-10$ and place it in the next free slot of the middle row.
$ \begin{array}{lllllll}
& & -- & -- & -- & -- & \\
-2 & ) & 1 & 7 & 13 & 6 & \\
& & & -2 & -10 & & \\
& & -- & -- & -- & -- & \\
& & 1 & 5 & & & \\
& & & & & &
\end{array}$
Add $13+(-10) =3$ and place 3 in the bottom row.
Multiply $(-2)(3)=-9$ and place it in the next free slot of the middle row.
$ \begin{array}{lllllll}
& & -- & -- & -- & -- & \\
-2 & ) & 1 & 7 & 13 & 6 & \\
& & & -2 & -10 & -6 & \\
& & -- & -- & -- & -- & \\
& & 1 & 5 & 3 & & \\
& & & & & &
\end{array}$
Add $6+(-6)=0$ and place it in the bottom row.
$ \begin{array}{lllllll}
& & -- & -- & -- & -- & \\
-2 & ) & 1 & 7 & 13 & 6 & \\
& & & -2 & -10 & -6 & \\
& & -- & -- & -- & -- & \\
& & 1 & 5 & 3 & 0 & \\
& & & & & &
\end{array}$
Interpret the result:
The last number of the bottom row represents the remainder, $0.$
The rest of the botom row holds coefficients of the quotient, which has a degree of one less than $f(x):\quad q(x)=x^{2}+5x+3$
So, $\displaystyle \quad \frac{f(x)}{x-k}=q(x)+\frac{r}{x-k}$
$\displaystyle \frac{x^{3}+7x^{2}+13x+6}{x+2}=\quad x^{2}+5x+3+\frac{0}{x+2}\quad =x^{2}+5x+3$
