The synthetic division table is:
$$ \begin{array}{c|rr}-3&-10&-14\\& & \color{black}{30} \\ \hline &\color{blue}{-10}&\color{orangered}{16} \end{array} $$The solution is:
$$ \frac{ -10x-14 }{ x+3 } = \color{blue}{-10} ~+~ \frac{ \color{red}{ 16 } }{ x+3 } $$Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 3 = 0 $ ( $ x = \color{blue}{ -3 } $ ) at the left.
$$ \begin{array}{c|rr}\color{blue}{-3}&-10&-14\\& & \\ \hline && \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rr}-3&\color{orangered}{ -10 }&-14\\& & \\ \hline &\color{orangered}{-10}& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -10 \right) } = \color{blue}{ 30 } $.
$$ \begin{array}{c|rr}\color{blue}{-3}&-10&-14\\& & \color{blue}{30} \\ \hline &\color{blue}{-10}& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -14 } + \color{orangered}{ 30 } = \color{orangered}{ 16 } $
$$ \begin{array}{c|rr}-3&-10&\color{orangered}{ -14 }\\& & \color{orangered}{30} \\ \hline &\color{blue}{-10}&\color{orangered}{16} \end{array} $$Bottom line represents the quotient $ \color{blue}{ -10 } $ with a remainder of $ \color{red}{ 16 } $.