Divide using synthetic division $$ ( 6x^{3}+9x^{2}-6x+7 ) \div ( x+1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-1&6&9&-6&7\\& & -6& -3& \color{black}{9} \\ \hline &\color{blue}{6}&\color{blue}{3}&\color{blue}{-9}&\color{orangered}{16} \end{array} $$The solution is:
$$ \frac{ 6x^{3}+9x^{2}-6x+7 }{ x+1 } = \color{blue}{6x^{2}+3x-9} ~+~ \frac{ \color{red}{ 16 } }{ x+1 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&6&9&-6&7\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-1&\color{orangered}{ 6 }&9&-6&7\\& & & & \\ \hline &\color{orangered}{6}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 6 } = \color{blue}{ -6 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&6&9&-6&7\\& & \color{blue}{-6} & & \\ \hline &\color{blue}{6}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 9 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrr}-1&6&\color{orangered}{ 9 }&-6&7\\& & \color{orangered}{-6} & & \\ \hline &6&\color{orangered}{3}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 3 } = \color{blue}{ -3 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&6&9&-6&7\\& & -6& \color{blue}{-3} & \\ \hline &6&\color{blue}{3}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ -9 } $
$$ \begin{array}{c|rrrr}-1&6&9&\color{orangered}{ -6 }&7\\& & -6& \color{orangered}{-3} & \\ \hline &6&3&\color{orangered}{-9}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ \left( -9 \right) } = \color{blue}{ 9 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&6&9&-6&7\\& & -6& -3& \color{blue}{9} \\ \hline &6&3&\color{blue}{-9}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ 9 } = \color{orangered}{ 16 } $
$$ \begin{array}{c|rrrr}-1&6&9&-6&\color{orangered}{ 7 }\\& & -6& -3& \color{orangered}{9} \\ \hline &\color{blue}{6}&\color{blue}{3}&\color{blue}{-9}&\color{orangered}{16} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 6x^{2}+3x-9 } $ with a remainder of $ \color{red}{ 16 } $.