The synthetic division table is:
$$ \begin{array}{c|rrrr}-2&5&-4&-31&-6\\& & -10& 28& \color{black}{6} \\ \hline &\color{blue}{5}&\color{blue}{-14}&\color{blue}{-3}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 5x^{3}-4x^{2}-31x-6 }{ x+2 } = \color{blue}{5x^{2}-14x-3} $$Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 2 = 0 $ ( $ x = \color{blue}{ -2 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&5&-4&-31&-6\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-2&\color{orangered}{ 5 }&-4&-31&-6\\& & & & \\ \hline &\color{orangered}{5}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 5 } = \color{blue}{ -10 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&5&-4&-31&-6\\& & \color{blue}{-10} & & \\ \hline &\color{blue}{5}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ \left( -10 \right) } = \color{orangered}{ -14 } $
$$ \begin{array}{c|rrrr}-2&5&\color{orangered}{ -4 }&-31&-6\\& & \color{orangered}{-10} & & \\ \hline &5&\color{orangered}{-14}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -14 \right) } = \color{blue}{ 28 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&5&-4&-31&-6\\& & -10& \color{blue}{28} & \\ \hline &5&\color{blue}{-14}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -31 } + \color{orangered}{ 28 } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrr}-2&5&-4&\color{orangered}{ -31 }&-6\\& & -10& \color{orangered}{28} & \\ \hline &5&-14&\color{orangered}{-3}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&5&-4&-31&-6\\& & -10& 28& \color{blue}{6} \\ \hline &5&-14&\color{blue}{-3}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ 6 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}-2&5&-4&-31&\color{orangered}{ -6 }\\& & -10& 28& \color{orangered}{6} \\ \hline &\color{blue}{5}&\color{blue}{-14}&\color{blue}{-3}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 5x^{2}-14x-3 } $ with a remainder of $ \color{red}{ 0 } $.