Divide using synthetic division $$ ( x^{5}-6x^{3}+18x+40 ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}2&1&0&-6&0&18&40\\& & 2& 4& -4& -8& \color{black}{20} \\ \hline &\color{blue}{1}&\color{blue}{2}&\color{blue}{-2}&\color{blue}{-4}&\color{blue}{10}&\color{orangered}{60} \end{array} $$The solution is:
$$ \frac{ x^{5}-6x^{3}+18x+40 }{ x-2 } = \color{blue}{x^{4}+2x^{3}-2x^{2}-4x+10} ~+~ \frac{ \color{red}{ 60 } }{ x-2 } $$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.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&1&0&-6&0&18&40\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}2&\color{orangered}{ 1 }&0&-6&0&18&40\\& & & & & & \\ \hline &\color{orangered}{1}&&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&1&0&-6&0&18&40\\& & \color{blue}{2} & & & & \\ \hline &\color{blue}{1}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 2 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrrr}2&1&\color{orangered}{ 0 }&-6&0&18&40\\& & \color{orangered}{2} & & & & \\ \hline &1&\color{orangered}{2}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 2 } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&1&0&-6&0&18&40\\& & 2& \color{blue}{4} & & & \\ \hline &1&\color{blue}{2}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ 4 } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrrrr}2&1&0&\color{orangered}{ -6 }&0&18&40\\& & 2& \color{orangered}{4} & & & \\ \hline &1&2&\color{orangered}{-2}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&1&0&-6&0&18&40\\& & 2& 4& \color{blue}{-4} & & \\ \hline &1&2&\color{blue}{-2}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrrrr}2&1&0&-6&\color{orangered}{ 0 }&18&40\\& & 2& 4& \color{orangered}{-4} & & \\ \hline &1&2&-2&\color{orangered}{-4}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ -8 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&1&0&-6&0&18&40\\& & 2& 4& -4& \color{blue}{-8} & \\ \hline &1&2&-2&\color{blue}{-4}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 18 } + \color{orangered}{ \left( -8 \right) } = \color{orangered}{ 10 } $
$$ \begin{array}{c|rrrrrr}2&1&0&-6&0&\color{orangered}{ 18 }&40\\& & 2& 4& -4& \color{orangered}{-8} & \\ \hline &1&2&-2&-4&\color{orangered}{10}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 10 } = \color{blue}{ 20 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&1&0&-6&0&18&40\\& & 2& 4& -4& -8& \color{blue}{20} \\ \hline &1&2&-2&-4&\color{blue}{10}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ 40 } + \color{orangered}{ 20 } = \color{orangered}{ 60 } $
$$ \begin{array}{c|rrrrrr}2&1&0&-6&0&18&\color{orangered}{ 40 }\\& & 2& 4& -4& -8& \color{orangered}{20} \\ \hline &\color{blue}{1}&\color{blue}{2}&\color{blue}{-2}&\color{blue}{-4}&\color{blue}{10}&\color{orangered}{60} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{4}+2x^{3}-2x^{2}-4x+10 } $ with a remainder of $ \color{red}{ 60 } $.