The synthetic division table is:
$$ \begin{array}{c|rrrr}-4&1&7&10&2\\& & -4& -12& \color{black}{8} \\ \hline &\color{blue}{1}&\color{blue}{3}&\color{blue}{-2}&\color{orangered}{10} \end{array} $$The solution is:
$$ \frac{ x^{3}+7x^{2}+10x+2 }{ x+4 } = \color{blue}{x^{2}+3x-2} ~+~ \frac{ \color{red}{ 10 } }{ x+4 } $$Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 4 = 0 $ ( $ x = \color{blue}{ -4 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&1&7&10&2\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-4&\color{orangered}{ 1 }&7&10&2\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 1 } = \color{blue}{ -4 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&1&7&10&2\\& & \color{blue}{-4} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrr}-4&1&\color{orangered}{ 7 }&10&2\\& & \color{orangered}{-4} & & \\ \hline &1&\color{orangered}{3}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 3 } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&1&7&10&2\\& & -4& \color{blue}{-12} & \\ \hline &1&\color{blue}{3}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 10 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrr}-4&1&7&\color{orangered}{ 10 }&2\\& & -4& \color{orangered}{-12} & \\ \hline &1&3&\color{orangered}{-2}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ 8 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&1&7&10&2\\& & -4& -12& \color{blue}{8} \\ \hline &1&3&\color{blue}{-2}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 8 } = \color{orangered}{ 10 } $
$$ \begin{array}{c|rrrr}-4&1&7&10&\color{orangered}{ 2 }\\& & -4& -12& \color{orangered}{8} \\ \hline &\color{blue}{1}&\color{blue}{3}&\color{blue}{-2}&\color{orangered}{10} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}+3x-2 } $ with a remainder of $ \color{red}{ 10 } $.