Divide using synthetic division $$ ( 10x^{4}+83x^{3}+21x^{2}-31x-62 ) \div ( x+8 ) $$
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-8&10&83&21&-31&-62\\& & -80& -24& 24& \color{black}{56} \\ \hline &\color{blue}{10}&\color{blue}{3}&\color{blue}{-3}&\color{blue}{-7}&\color{orangered}{-6} \end{array} $$The solution is:
$$ \dfrac{ 10x^{4}+83x^{3}+21x^{2}-31x-62 }{ x+8 } = \color{blue}{10x^{3}+3x^{2}-3x-7} \color{red}{~-~} \dfrac{ \color{red}{ 6 } }{ x+8 } $$Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 8 = 0 $ ( $ x = \color{blue}{ -8 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{-8}&10&83&21&-31&-62\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-8&\color{orangered}{ 10 }&83&21&-31&-62\\& & & & & \\ \hline &\color{orangered}{10}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -8 } \cdot \color{blue}{ 10 } = \color{blue}{ -80 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-8}&10&83&21&-31&-62\\& & \color{blue}{-80} & & & \\ \hline &\color{blue}{10}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 83 } + \color{orangered}{ \left( -80 \right) } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrrr}-8&10&\color{orangered}{ 83 }&21&-31&-62\\& & \color{orangered}{-80} & & & \\ \hline &10&\color{orangered}{3}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -8 } \cdot \color{blue}{ 3 } = \color{blue}{ -24 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-8}&10&83&21&-31&-62\\& & -80& \color{blue}{-24} & & \\ \hline &10&\color{blue}{3}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 21 } + \color{orangered}{ \left( -24 \right) } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrrr}-8&10&83&\color{orangered}{ 21 }&-31&-62\\& & -80& \color{orangered}{-24} & & \\ \hline &10&3&\color{orangered}{-3}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -8 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ 24 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-8}&10&83&21&-31&-62\\& & -80& -24& \color{blue}{24} & \\ \hline &10&3&\color{blue}{-3}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -31 } + \color{orangered}{ 24 } = \color{orangered}{ -7 } $
$$ \begin{array}{c|rrrrr}-8&10&83&21&\color{orangered}{ -31 }&-62\\& & -80& -24& \color{orangered}{24} & \\ \hline &10&3&-3&\color{orangered}{-7}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -8 } \cdot \color{blue}{ \left( -7 \right) } = \color{blue}{ 56 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-8}&10&83&21&-31&-62\\& & -80& -24& 24& \color{blue}{56} \\ \hline &10&3&-3&\color{blue}{-7}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -62 } + \color{orangered}{ 56 } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrrr}-8&10&83&21&-31&\color{orangered}{ -62 }\\& & -80& -24& 24& \color{orangered}{56} \\ \hline &\color{blue}{10}&\color{blue}{3}&\color{blue}{-3}&\color{blue}{-7}&\color{orangered}{-6} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 10x^{3}+3x^{2}-3x-7 } $ with a remainder of $ \color{red}{ -6 } $.