Divide using synthetic division $$ ( 2x^{5}+4x^{4}-2x^{3}+3x^{2}+3x+3 ) \div ( x-1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}1&2&4&-2&3&3&3\\& & 2& 6& 4& 7& \color{black}{10} \\ \hline &\color{blue}{2}&\color{blue}{6}&\color{blue}{4}&\color{blue}{7}&\color{blue}{10}&\color{orangered}{13} \end{array} $$The solution is:
$$ \frac{ 2x^{5}+4x^{4}-2x^{3}+3x^{2}+3x+3 }{ x-1 } = \color{blue}{2x^{4}+6x^{3}+4x^{2}+7x+10} ~+~ \frac{ \color{red}{ 13 } }{ x-1 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -1 = 0 $ ( $ x = \color{blue}{ 1 } $ ) at the left.
$$ \begin{array}{c|rrrrrr}\color{blue}{1}&2&4&-2&3&3&3\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}1&\color{orangered}{ 2 }&4&-2&3&3&3\\& & & & & & \\ \hline &\color{orangered}{2}&&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 2 } = \color{blue}{ 2 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{1}&2&4&-2&3&3&3\\& & \color{blue}{2} & & & & \\ \hline &\color{blue}{2}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ 2 } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrrrr}1&2&\color{orangered}{ 4 }&-2&3&3&3\\& & \color{orangered}{2} & & & & \\ \hline &2&\color{orangered}{6}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 6 } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{1}&2&4&-2&3&3&3\\& & 2& \color{blue}{6} & & & \\ \hline &2&\color{blue}{6}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ 6 } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrrrr}1&2&4&\color{orangered}{ -2 }&3&3&3\\& & 2& \color{orangered}{6} & & & \\ \hline &2&6&\color{orangered}{4}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 4 } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{1}&2&4&-2&3&3&3\\& & 2& 6& \color{blue}{4} & & \\ \hline &2&6&\color{blue}{4}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 4 } = \color{orangered}{ 7 } $
$$ \begin{array}{c|rrrrrr}1&2&4&-2&\color{orangered}{ 3 }&3&3\\& & 2& 6& \color{orangered}{4} & & \\ \hline &2&6&4&\color{orangered}{7}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 7 } = \color{blue}{ 7 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{1}&2&4&-2&3&3&3\\& & 2& 6& 4& \color{blue}{7} & \\ \hline &2&6&4&\color{blue}{7}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 7 } = \color{orangered}{ 10 } $
$$ \begin{array}{c|rrrrrr}1&2&4&-2&3&\color{orangered}{ 3 }&3\\& & 2& 6& 4& \color{orangered}{7} & \\ \hline &2&6&4&7&\color{orangered}{10}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 10 } = \color{blue}{ 10 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{1}&2&4&-2&3&3&3\\& & 2& 6& 4& 7& \color{blue}{10} \\ \hline &2&6&4&7&\color{blue}{10}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 10 } = \color{orangered}{ 13 } $
$$ \begin{array}{c|rrrrrr}1&2&4&-2&3&3&\color{orangered}{ 3 }\\& & 2& 6& 4& 7& \color{orangered}{10} \\ \hline &\color{blue}{2}&\color{blue}{6}&\color{blue}{4}&\color{blue}{7}&\color{blue}{10}&\color{orangered}{13} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{4}+6x^{3}+4x^{2}+7x+10 } $ with a remainder of $ \color{red}{ 13 } $.