Divide using synthetic division $$ ( x^{4}-3x^{3}-6x+8 ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}2&1&-3&0&-6&8\\& & 2& -2& -4& \color{black}{-20} \\ \hline &\color{blue}{1}&\color{blue}{-1}&\color{blue}{-2}&\color{blue}{-10}&\color{orangered}{-12} \end{array} $$The solution is:
$$ \frac{ x^{4}-3x^{3}-6x+8 }{ x-2 } = \color{blue}{x^{3}-x^{2}-2x-10} \color{red}{~-~} \frac{ \color{red}{ 12 } }{ x-2 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&-3&0&-6&8\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}2&\color{orangered}{ 1 }&-3&0&-6&8\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 1 } = \color{blue}{ 2 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&-3&0&-6&8\\& & \color{blue}{2} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ 2 } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrrr}2&1&\color{orangered}{ -3 }&0&-6&8\\& & \color{orangered}{2} & & & \\ \hline &1&\color{orangered}{-1}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ -2 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&-3&0&-6&8\\& & 2& \color{blue}{-2} & & \\ \hline &1&\color{blue}{-1}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -2 \right) } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrrr}2&1&-3&\color{orangered}{ 0 }&-6&8\\& & 2& \color{orangered}{-2} & & \\ \hline &1&-1&\color{orangered}{-2}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ -4 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&-3&0&-6&8\\& & 2& -2& \color{blue}{-4} & \\ \hline &1&-1&\color{blue}{-2}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ -10 } $
$$ \begin{array}{c|rrrrr}2&1&-3&0&\color{orangered}{ -6 }&8\\& & 2& -2& \color{orangered}{-4} & \\ \hline &1&-1&-2&\color{orangered}{-10}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -10 \right) } = \color{blue}{ -20 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&-3&0&-6&8\\& & 2& -2& -4& \color{blue}{-20} \\ \hline &1&-1&-2&\color{blue}{-10}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 8 } + \color{orangered}{ \left( -20 \right) } = \color{orangered}{ -12 } $
$$ \begin{array}{c|rrrrr}2&1&-3&0&-6&\color{orangered}{ 8 }\\& & 2& -2& -4& \color{orangered}{-20} \\ \hline &\color{blue}{1}&\color{blue}{-1}&\color{blue}{-2}&\color{blue}{-10}&\color{orangered}{-12} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}-x^{2}-2x-10 } $ with a remainder of $ \color{red}{ -12 } $.