Divide using synthetic division $$ ( 2x^{5}+x^{4}+6x^{3}+x^{2}-7x-4 ) \div ( 2x+1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}-1&2&1&6&1&-7&-4\\& & -2& 1& -7& 6& \color{black}{1} \\ \hline &\color{blue}{2}&\color{blue}{-1}&\color{blue}{7}&\color{blue}{-6}&\color{blue}{-1}&\color{orangered}{-3} \end{array} $$The solution is:
$$ \frac{ 2x^{5}+x^{4}+6x^{3}+x^{2}-7x-4 }{ x+1 } = \color{blue}{2x^{4}-x^{3}+7x^{2}-6x-1} \color{red}{~-~} \frac{ \color{red}{ 3 } }{ x+1 } $$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.
$$ \begin{array}{c|rrrrrr}\color{blue}{-1}&2&1&6&1&-7&-4\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}-1&\color{orangered}{ 2 }&1&6&1&-7&-4\\& & & & & & \\ \hline &\color{orangered}{2}&&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-1}&2&1&6&1&-7&-4\\& & \color{blue}{-2} & & & & \\ \hline &\color{blue}{2}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ \left( -2 \right) } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrrrr}-1&2&\color{orangered}{ 1 }&6&1&-7&-4\\& & \color{orangered}{-2} & & & & \\ \hline &2&\color{orangered}{-1}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ 1 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-1}&2&1&6&1&-7&-4\\& & -2& \color{blue}{1} & & & \\ \hline &2&\color{blue}{-1}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ 1 } = \color{orangered}{ 7 } $
$$ \begin{array}{c|rrrrrr}-1&2&1&\color{orangered}{ 6 }&1&-7&-4\\& & -2& \color{orangered}{1} & & & \\ \hline &2&-1&\color{orangered}{7}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 7 } = \color{blue}{ -7 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-1}&2&1&6&1&-7&-4\\& & -2& 1& \color{blue}{-7} & & \\ \hline &2&-1&\color{blue}{7}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ \left( -7 \right) } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrrrr}-1&2&1&6&\color{orangered}{ 1 }&-7&-4\\& & -2& 1& \color{orangered}{-7} & & \\ \hline &2&-1&7&\color{orangered}{-6}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ \left( -6 \right) } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-1}&2&1&6&1&-7&-4\\& & -2& 1& -7& \color{blue}{6} & \\ \hline &2&-1&7&\color{blue}{-6}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 6 } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrrrr}-1&2&1&6&1&\color{orangered}{ -7 }&-4\\& & -2& 1& -7& \color{orangered}{6} & \\ \hline &2&-1&7&-6&\color{orangered}{-1}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ 1 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-1}&2&1&6&1&-7&-4\\& & -2& 1& -7& 6& \color{blue}{1} \\ \hline &2&-1&7&-6&\color{blue}{-1}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ 1 } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrrrr}-1&2&1&6&1&-7&\color{orangered}{ -4 }\\& & -2& 1& -7& 6& \color{orangered}{1} \\ \hline &\color{blue}{2}&\color{blue}{-1}&\color{blue}{7}&\color{blue}{-6}&\color{blue}{-1}&\color{orangered}{-3} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{4}-x^{3}+7x^{2}-6x-1 } $ with a remainder of $ \color{red}{ -3 } $.