Divide using synthetic division $$ ( x^{3}-x^{2}-10x-8 ) \div ( x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-3&1&-1&-10&-8\\& & -3& 12& \color{black}{-6} \\ \hline &\color{blue}{1}&\color{blue}{-4}&\color{blue}{2}&\color{orangered}{-14} \end{array} $$The solution is:
$$ \frac{ x^{3}-x^{2}-10x-8 }{ x+3 } = \color{blue}{x^{2}-4x+2} \color{red}{~-~} \frac{ \color{red}{ 14 } }{ x+3 } $$Explanation
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.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&1&-1&-10&-8\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-3&\color{orangered}{ 1 }&-1&-10&-8\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 1 } = \color{blue}{ -3 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&1&-1&-10&-8\\& & \color{blue}{-3} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrr}-3&1&\color{orangered}{ -1 }&-10&-8\\& & \color{orangered}{-3} & & \\ \hline &1&\color{orangered}{-4}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&1&-1&-10&-8\\& & -3& \color{blue}{12} & \\ \hline &1&\color{blue}{-4}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -10 } + \color{orangered}{ 12 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrr}-3&1&-1&\color{orangered}{ -10 }&-8\\& & -3& \color{orangered}{12} & \\ \hline &1&-4&\color{orangered}{2}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 2 } = \color{blue}{ -6 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&1&-1&-10&-8\\& & -3& 12& \color{blue}{-6} \\ \hline &1&-4&\color{blue}{2}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -8 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ -14 } $
$$ \begin{array}{c|rrrr}-3&1&-1&-10&\color{orangered}{ -8 }\\& & -3& 12& \color{orangered}{-6} \\ \hline &\color{blue}{1}&\color{blue}{-4}&\color{blue}{2}&\color{orangered}{-14} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}-4x+2 } $ with a remainder of $ \color{red}{ -14 } $.