Divide using synthetic division $$ ( x^{3}-8x^{2}+6x-11 ) \div ( x+1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-1&1&-8&6&-11\\& & -1& 9& \color{black}{-15} \\ \hline &\color{blue}{1}&\color{blue}{-9}&\color{blue}{15}&\color{orangered}{-26} \end{array} $$The solution is:
$$ \frac{ x^{3}-8x^{2}+6x-11 }{ x+1 } = \color{blue}{x^{2}-9x+15} \color{red}{~-~} \frac{ \color{red}{ 26 } }{ x+1 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 1 = 0 $ ( $ x = \color{blue}{ -1 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&1&-8&6&-11\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-1&\color{orangered}{ 1 }&-8&6&-11\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 1 } = \color{blue}{ -1 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&1&-8&6&-11\\& & \color{blue}{-1} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -8 } + \color{orangered}{ \left( -1 \right) } = \color{orangered}{ -9 } $
$$ \begin{array}{c|rrrr}-1&1&\color{orangered}{ -8 }&6&-11\\& & \color{orangered}{-1} & & \\ \hline &1&\color{orangered}{-9}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ \left( -9 \right) } = \color{blue}{ 9 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&1&-8&6&-11\\& & -1& \color{blue}{9} & \\ \hline &1&\color{blue}{-9}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ 9 } = \color{orangered}{ 15 } $
$$ \begin{array}{c|rrrr}-1&1&-8&\color{orangered}{ 6 }&-11\\& & -1& \color{orangered}{9} & \\ \hline &1&-9&\color{orangered}{15}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 15 } = \color{blue}{ -15 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&1&-8&6&-11\\& & -1& 9& \color{blue}{-15} \\ \hline &1&-9&\color{blue}{15}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -11 } + \color{orangered}{ \left( -15 \right) } = \color{orangered}{ -26 } $
$$ \begin{array}{c|rrrr}-1&1&-8&6&\color{orangered}{ -11 }\\& & -1& 9& \color{orangered}{-15} \\ \hline &\color{blue}{1}&\color{blue}{-9}&\color{blue}{15}&\color{orangered}{-26} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}-9x+15 } $ with a remainder of $ \color{red}{ -26 } $.