Divide using synthetic division $$ ( 6x^{4}-x^{3}-32x^{2}+5x+10 ) \div ( 2x+1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-1&6&-1&-32&5&10\\& & -6& 7& 25& \color{black}{-30} \\ \hline &\color{blue}{6}&\color{blue}{-7}&\color{blue}{-25}&\color{blue}{30}&\color{orangered}{-20} \end{array} $$The solution is:
$$ \frac{ 6x^{4}-x^{3}-32x^{2}+5x+10 }{ x+1 } = \color{blue}{6x^{3}-7x^{2}-25x+30} \color{red}{~-~} \frac{ \color{red}{ 20 } }{ x+1 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&6&-1&-32&5&10\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-1&\color{orangered}{ 6 }&-1&-32&5&10\\& & & & & \\ \hline &\color{orangered}{6}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&6&-1&-32&5&10\\& & \color{blue}{-6} & & & \\ \hline &\color{blue}{6}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ -7 } $
$$ \begin{array}{c|rrrrr}-1&6&\color{orangered}{ -1 }&-32&5&10\\& & \color{orangered}{-6} & & & \\ \hline &6&\color{orangered}{-7}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ \left( -7 \right) } = \color{blue}{ 7 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&6&-1&-32&5&10\\& & -6& \color{blue}{7} & & \\ \hline &6&\color{blue}{-7}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -32 } + \color{orangered}{ 7 } = \color{orangered}{ -25 } $
$$ \begin{array}{c|rrrrr}-1&6&-1&\color{orangered}{ -32 }&5&10\\& & -6& \color{orangered}{7} & & \\ \hline &6&-7&\color{orangered}{-25}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ \left( -25 \right) } = \color{blue}{ 25 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&6&-1&-32&5&10\\& & -6& 7& \color{blue}{25} & \\ \hline &6&-7&\color{blue}{-25}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ 25 } = \color{orangered}{ 30 } $
$$ \begin{array}{c|rrrrr}-1&6&-1&-32&\color{orangered}{ 5 }&10\\& & -6& 7& \color{orangered}{25} & \\ \hline &6&-7&-25&\color{orangered}{30}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 30 } = \color{blue}{ -30 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&6&-1&-32&5&10\\& & -6& 7& 25& \color{blue}{-30} \\ \hline &6&-7&-25&\color{blue}{30}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 10 } + \color{orangered}{ \left( -30 \right) } = \color{orangered}{ -20 } $
$$ \begin{array}{c|rrrrr}-1&6&-1&-32&5&\color{orangered}{ 10 }\\& & -6& 7& 25& \color{orangered}{-30} \\ \hline &\color{blue}{6}&\color{blue}{-7}&\color{blue}{-25}&\color{blue}{30}&\color{orangered}{-20} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 6x^{3}-7x^{2}-25x+30 } $ with a remainder of $ \color{red}{ -20 } $.