Divide using synthetic division $$ ( x^{5}+x^{3}-x^{2} ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}2&1&0&1&-1&0&0\\& & 2& 4& 10& 18& \color{black}{36} \\ \hline &\color{blue}{1}&\color{blue}{2}&\color{blue}{5}&\color{blue}{9}&\color{blue}{18}&\color{orangered}{36} \end{array} $$The solution is:
$$ \frac{ x^{5}+x^{3}-x^{2} }{ x-2 } = \color{blue}{x^{4}+2x^{3}+5x^{2}+9x+18} ~+~ \frac{ \color{red}{ 36 } }{ x-2 } $$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.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&1&0&1&-1&0&0\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}2&\color{orangered}{ 1 }&0&1&-1&0&0\\& & & & & & \\ \hline &\color{orangered}{1}&&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&1&0&1&-1&0&0\\& & \color{blue}{2} & & & & \\ \hline &\color{blue}{1}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 2 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrrr}2&1&\color{orangered}{ 0 }&1&-1&0&0\\& & \color{orangered}{2} & & & & \\ \hline &1&\color{orangered}{2}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&1&0&1&-1&0&0\\& & 2& \color{blue}{4} & & & \\ \hline &1&\color{blue}{2}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 4 } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrrrr}2&1&0&\color{orangered}{ 1 }&-1&0&0\\& & 2& \color{orangered}{4} & & & \\ \hline &1&2&\color{orangered}{5}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 5 } = \color{blue}{ 10 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&1&0&1&-1&0&0\\& & 2& 4& \color{blue}{10} & & \\ \hline &1&2&\color{blue}{5}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 10 } = \color{orangered}{ 9 } $
$$ \begin{array}{c|rrrrrr}2&1&0&1&\color{orangered}{ -1 }&0&0\\& & 2& 4& \color{orangered}{10} & & \\ \hline &1&2&5&\color{orangered}{9}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 9 } = \color{blue}{ 18 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&1&0&1&-1&0&0\\& & 2& 4& 10& \color{blue}{18} & \\ \hline &1&2&5&\color{blue}{9}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 18 } = \color{orangered}{ 18 } $
$$ \begin{array}{c|rrrrrr}2&1&0&1&-1&\color{orangered}{ 0 }&0\\& & 2& 4& 10& \color{orangered}{18} & \\ \hline &1&2&5&9&\color{orangered}{18}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 18 } = \color{blue}{ 36 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&1&0&1&-1&0&0\\& & 2& 4& 10& 18& \color{blue}{36} \\ \hline &1&2&5&9&\color{blue}{18}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 36 } = \color{orangered}{ 36 } $
$$ \begin{array}{c|rrrrrr}2&1&0&1&-1&0&\color{orangered}{ 0 }\\& & 2& 4& 10& 18& \color{orangered}{36} \\ \hline &\color{blue}{1}&\color{blue}{2}&\color{blue}{5}&\color{blue}{9}&\color{blue}{18}&\color{orangered}{36} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{4}+2x^{3}+5x^{2}+9x+18 } $ with a remainder of $ \color{red}{ 36 } $.