Divide using synthetic division $$ ( 2x^{4}-x^{3}-7x^{2}+9x+6 ) \div ( 2x+1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-1&2&-1&-7&9&6\\& & -2& 3& 4& \color{black}{-13} \\ \hline &\color{blue}{2}&\color{blue}{-3}&\color{blue}{-4}&\color{blue}{13}&\color{orangered}{-7} \end{array} $$The solution is:
$$ \frac{ 2x^{4}-x^{3}-7x^{2}+9x+6 }{ x+1 } = \color{blue}{2x^{3}-3x^{2}-4x+13} \color{red}{~-~} \frac{ \color{red}{ 7 } }{ x+1 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&2&-1&-7&9&6\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-1&\color{orangered}{ 2 }&-1&-7&9&6\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 2 } = \color{blue}{ -2 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&2&-1&-7&9&6\\& & \color{blue}{-2} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ \left( -2 \right) } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrrr}-1&2&\color{orangered}{ -1 }&-7&9&6\\& & \color{orangered}{-2} & & & \\ \hline &2&\color{orangered}{-3}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&2&-1&-7&9&6\\& & -2& \color{blue}{3} & & \\ \hline &2&\color{blue}{-3}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 3 } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrrr}-1&2&-1&\color{orangered}{ -7 }&9&6\\& & -2& \color{orangered}{3} & & \\ \hline &2&-3&\color{orangered}{-4}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&2&-1&-7&9&6\\& & -2& 3& \color{blue}{4} & \\ \hline &2&-3&\color{blue}{-4}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 9 } + \color{orangered}{ 4 } = \color{orangered}{ 13 } $
$$ \begin{array}{c|rrrrr}-1&2&-1&-7&\color{orangered}{ 9 }&6\\& & -2& 3& \color{orangered}{4} & \\ \hline &2&-3&-4&\color{orangered}{13}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 13 } = \color{blue}{ -13 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&2&-1&-7&9&6\\& & -2& 3& 4& \color{blue}{-13} \\ \hline &2&-3&-4&\color{blue}{13}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ \left( -13 \right) } = \color{orangered}{ -7 } $
$$ \begin{array}{c|rrrrr}-1&2&-1&-7&9&\color{orangered}{ 6 }\\& & -2& 3& 4& \color{orangered}{-13} \\ \hline &\color{blue}{2}&\color{blue}{-3}&\color{blue}{-4}&\color{blue}{13}&\color{orangered}{-7} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{3}-3x^{2}-4x+13 } $ with a remainder of $ \color{red}{ -7 } $.