Divide using synthetic division $$ ( 6x^{4}-x^{3}-32x^{2}+5x+10 ) \div ( 2x-1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}1&6&-1&-32&5&10\\& & 6& 5& -27& \color{black}{-22} \\ \hline &\color{blue}{6}&\color{blue}{5}&\color{blue}{-27}&\color{blue}{-22}&\color{orangered}{-12} \end{array} $$The solution is:
$$ \frac{ 6x^{4}-x^{3}-32x^{2}+5x+10 }{ x-1 } = \color{blue}{6x^{3}+5x^{2}-27x-22} \color{red}{~-~} \frac{ \color{red}{ 12 } }{ x-1 } $$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}{ 6 } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrrr}1&6&\color{orangered}{ -1 }&-32&5&10\\& & \color{orangered}{6} & & & \\ \hline &6&\color{orangered}{5}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 5 } = \color{blue}{ 5 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&6&-1&-32&5&10\\& & 6& \color{blue}{5} & & \\ \hline &6&\color{blue}{5}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -32 } + \color{orangered}{ 5 } = \color{orangered}{ -27 } $
$$ \begin{array}{c|rrrrr}1&6&-1&\color{orangered}{ -32 }&5&10\\& & 6& \color{orangered}{5} & & \\ \hline &6&5&\color{orangered}{-27}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -27 \right) } = \color{blue}{ -27 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&6&-1&-32&5&10\\& & 6& 5& \color{blue}{-27} & \\ \hline &6&5&\color{blue}{-27}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ \left( -27 \right) } = \color{orangered}{ -22 } $
$$ \begin{array}{c|rrrrr}1&6&-1&-32&\color{orangered}{ 5 }&10\\& & 6& 5& \color{orangered}{-27} & \\ \hline &6&5&-27&\color{orangered}{-22}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -22 \right) } = \color{blue}{ -22 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&6&-1&-32&5&10\\& & 6& 5& -27& \color{blue}{-22} \\ \hline &6&5&-27&\color{blue}{-22}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 10 } + \color{orangered}{ \left( -22 \right) } = \color{orangered}{ -12 } $
$$ \begin{array}{c|rrrrr}1&6&-1&-32&5&\color{orangered}{ 10 }\\& & 6& 5& -27& \color{orangered}{-22} \\ \hline &\color{blue}{6}&\color{blue}{5}&\color{blue}{-27}&\color{blue}{-22}&\color{orangered}{-12} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 6x^{3}+5x^{2}-27x-22 } $ with a remainder of $ \color{red}{ -12 } $.