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