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