Divide using synthetic division $$ ( 6x^{4}-x^{3}-32x^{2}+5x+10 ) \div ( 3x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-2&6&-1&-32&5&10\\& & -12& 26& 12& \color{black}{-34} \\ \hline &\color{blue}{6}&\color{blue}{-13}&\color{blue}{-6}&\color{blue}{17}&\color{orangered}{-24} \end{array} $$The solution is:
$$ \frac{ 6x^{4}-x^{3}-32x^{2}+5x+10 }{ x+2 } = \color{blue}{6x^{3}-13x^{2}-6x+17} \color{red}{~-~} \frac{ \color{red}{ 24 } }{ x+2 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 2 = 0 $ ( $ x = \color{blue}{ -2 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&6&-1&-32&5&10\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-2&\color{orangered}{ 6 }&-1&-32&5&10\\& & & & & \\ \hline &\color{orangered}{6}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 6 } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&6&-1&-32&5&10\\& & \color{blue}{-12} & & & \\ \hline &\color{blue}{6}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ -13 } $
$$ \begin{array}{c|rrrrr}-2&6&\color{orangered}{ -1 }&-32&5&10\\& & \color{orangered}{-12} & & & \\ \hline &6&\color{orangered}{-13}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -13 \right) } = \color{blue}{ 26 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&6&-1&-32&5&10\\& & -12& \color{blue}{26} & & \\ \hline &6&\color{blue}{-13}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -32 } + \color{orangered}{ 26 } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrrr}-2&6&-1&\color{orangered}{ -32 }&5&10\\& & -12& \color{orangered}{26} & & \\ \hline &6&-13&\color{orangered}{-6}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -6 \right) } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&6&-1&-32&5&10\\& & -12& 26& \color{blue}{12} & \\ \hline &6&-13&\color{blue}{-6}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ 12 } = \color{orangered}{ 17 } $
$$ \begin{array}{c|rrrrr}-2&6&-1&-32&\color{orangered}{ 5 }&10\\& & -12& 26& \color{orangered}{12} & \\ \hline &6&-13&-6&\color{orangered}{17}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 17 } = \color{blue}{ -34 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&6&-1&-32&5&10\\& & -12& 26& 12& \color{blue}{-34} \\ \hline &6&-13&-6&\color{blue}{17}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 10 } + \color{orangered}{ \left( -34 \right) } = \color{orangered}{ -24 } $
$$ \begin{array}{c|rrrrr}-2&6&-1&-32&5&\color{orangered}{ 10 }\\& & -12& 26& 12& \color{orangered}{-34} \\ \hline &\color{blue}{6}&\color{blue}{-13}&\color{blue}{-6}&\color{blue}{17}&\color{orangered}{-24} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 6x^{3}-13x^{2}-6x+17 } $ with a remainder of $ \color{red}{ -24 } $.