Divide using synthetic division $$ ( x^{3}-4x^{2}-7x-10 ) \div ( x+5 ) $$
The synthetic division table is:
$$ \begin{array}{c|rrrr}-5&1&-4&-7&-10\\& & -5& 45& \color{black}{-190} \\ \hline &\color{blue}{1}&\color{blue}{-9}&\color{blue}{38}&\color{orangered}{-200} \end{array} $$The solution is:
$$ \dfrac{ x^{3}-4x^{2}-7x-10 }{ x+5 } = \color{blue}{x^{2}-9x+38} \color{red}{~-~} \dfrac{ \color{red}{ 200 } }{ x+5 } $$Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 5 = 0 $ ( $ x = \color{blue}{ -5 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&1&-4&-7&-10\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-5&\color{orangered}{ 1 }&-4&-7&-10\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 1 } = \color{blue}{ -5 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&1&-4&-7&-10\\& & \color{blue}{-5} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ \left( -5 \right) } = \color{orangered}{ -9 } $
$$ \begin{array}{c|rrrr}-5&1&\color{orangered}{ -4 }&-7&-10\\& & \color{orangered}{-5} & & \\ \hline &1&\color{orangered}{-9}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ \left( -9 \right) } = \color{blue}{ 45 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&1&-4&-7&-10\\& & -5& \color{blue}{45} & \\ \hline &1&\color{blue}{-9}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 45 } = \color{orangered}{ 38 } $
$$ \begin{array}{c|rrrr}-5&1&-4&\color{orangered}{ -7 }&-10\\& & -5& \color{orangered}{45} & \\ \hline &1&-9&\color{orangered}{38}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 38 } = \color{blue}{ -190 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&1&-4&-7&-10\\& & -5& 45& \color{blue}{-190} \\ \hline &1&-9&\color{blue}{38}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -10 } + \color{orangered}{ \left( -190 \right) } = \color{orangered}{ -200 } $
$$ \begin{array}{c|rrrr}-5&1&-4&-7&\color{orangered}{ -10 }\\& & -5& 45& \color{orangered}{-190} \\ \hline &\color{blue}{1}&\color{blue}{-9}&\color{blue}{38}&\color{orangered}{-200} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}-9x+38 } $ with a remainder of $ \color{red}{ -200 } $.