Divide using synthetic division $$ ( x^{5}+5x^{3}-3x+7 ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}2&1&0&5&0&-3&7\\& & 2& 4& 18& 36& \color{black}{66} \\ \hline &\color{blue}{1}&\color{blue}{2}&\color{blue}{9}&\color{blue}{18}&\color{blue}{33}&\color{orangered}{73} \end{array} $$The solution is:
$$ \frac{ x^{5}+5x^{3}-3x+7 }{ x-2 } = \color{blue}{x^{4}+2x^{3}+9x^{2}+18x+33} ~+~ \frac{ \color{red}{ 73 } }{ x-2 } $$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.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&1&0&5&0&-3&7\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}2&\color{orangered}{ 1 }&0&5&0&-3&7\\& & & & & & \\ \hline &\color{orangered}{1}&&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&1&0&5&0&-3&7\\& & \color{blue}{2} & & & & \\ \hline &\color{blue}{1}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 2 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrrr}2&1&\color{orangered}{ 0 }&5&0&-3&7\\& & \color{orangered}{2} & & & & \\ \hline &1&\color{orangered}{2}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&1&0&5&0&-3&7\\& & 2& \color{blue}{4} & & & \\ \hline &1&\color{blue}{2}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ 4 } = \color{orangered}{ 9 } $
$$ \begin{array}{c|rrrrrr}2&1&0&\color{orangered}{ 5 }&0&-3&7\\& & 2& \color{orangered}{4} & & & \\ \hline &1&2&\color{orangered}{9}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 9 } = \color{blue}{ 18 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&1&0&5&0&-3&7\\& & 2& 4& \color{blue}{18} & & \\ \hline &1&2&\color{blue}{9}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 18 } = \color{orangered}{ 18 } $
$$ \begin{array}{c|rrrrrr}2&1&0&5&\color{orangered}{ 0 }&-3&7\\& & 2& 4& \color{orangered}{18} & & \\ \hline &1&2&9&\color{orangered}{18}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 18 } = \color{blue}{ 36 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&1&0&5&0&-3&7\\& & 2& 4& 18& \color{blue}{36} & \\ \hline &1&2&9&\color{blue}{18}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ 36 } = \color{orangered}{ 33 } $
$$ \begin{array}{c|rrrrrr}2&1&0&5&0&\color{orangered}{ -3 }&7\\& & 2& 4& 18& \color{orangered}{36} & \\ \hline &1&2&9&18&\color{orangered}{33}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 33 } = \color{blue}{ 66 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&1&0&5&0&-3&7\\& & 2& 4& 18& 36& \color{blue}{66} \\ \hline &1&2&9&18&\color{blue}{33}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ 66 } = \color{orangered}{ 73 } $
$$ \begin{array}{c|rrrrrr}2&1&0&5&0&-3&\color{orangered}{ 7 }\\& & 2& 4& 18& 36& \color{orangered}{66} \\ \hline &\color{blue}{1}&\color{blue}{2}&\color{blue}{9}&\color{blue}{18}&\color{blue}{33}&\color{orangered}{73} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{4}+2x^{3}+9x^{2}+18x+33 } $ with a remainder of $ \color{red}{ 73 } $.