Divide using synthetic division $$ ( 6x^{4}-x^{3}-32x^{2}+5x+10 ) \div ( 3x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}2&6&-1&-32&5&10\\& & 12& 22& -20& \color{black}{-30} \\ \hline &\color{blue}{6}&\color{blue}{11}&\color{blue}{-10}&\color{blue}{-15}&\color{orangered}{-20} \end{array} $$The solution is:
$$ \frac{ 6x^{4}-x^{3}-32x^{2}+5x+10 }{ x-2 } = \color{blue}{6x^{3}+11x^{2}-10x-15} \color{red}{~-~} \frac{ \color{red}{ 20 } }{ x-2 } $$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}{ 12 } = \color{orangered}{ 11 } $
$$ \begin{array}{c|rrrrr}2&6&\color{orangered}{ -1 }&-32&5&10\\& & \color{orangered}{12} & & & \\ \hline &6&\color{orangered}{11}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 11 } = \color{blue}{ 22 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&6&-1&-32&5&10\\& & 12& \color{blue}{22} & & \\ \hline &6&\color{blue}{11}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -32 } + \color{orangered}{ 22 } = \color{orangered}{ -10 } $
$$ \begin{array}{c|rrrrr}2&6&-1&\color{orangered}{ -32 }&5&10\\& & 12& \color{orangered}{22} & & \\ \hline &6&11&\color{orangered}{-10}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -10 \right) } = \color{blue}{ -20 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&6&-1&-32&5&10\\& & 12& 22& \color{blue}{-20} & \\ \hline &6&11&\color{blue}{-10}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ \left( -20 \right) } = \color{orangered}{ -15 } $
$$ \begin{array}{c|rrrrr}2&6&-1&-32&\color{orangered}{ 5 }&10\\& & 12& 22& \color{orangered}{-20} & \\ \hline &6&11&-10&\color{orangered}{-15}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -15 \right) } = \color{blue}{ -30 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&6&-1&-32&5&10\\& & 12& 22& -20& \color{blue}{-30} \\ \hline &6&11&-10&\color{blue}{-15}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 10 } + \color{orangered}{ \left( -30 \right) } = \color{orangered}{ -20 } $
$$ \begin{array}{c|rrrrr}2&6&-1&-32&5&\color{orangered}{ 10 }\\& & 12& 22& -20& \color{orangered}{-30} \\ \hline &\color{blue}{6}&\color{blue}{11}&\color{blue}{-10}&\color{blue}{-15}&\color{orangered}{-20} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 6x^{3}+11x^{2}-10x-15 } $ with a remainder of $ \color{red}{ -20 } $.