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