Divide using synthetic division $$ ( x^{4}-x^{3}+x^{2}-x+2 ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}2&1&-1&1&-1&2\\& & 2& 2& 6& \color{black}{10} \\ \hline &\color{blue}{1}&\color{blue}{1}&\color{blue}{3}&\color{blue}{5}&\color{orangered}{12} \end{array} $$The solution is:
$$ \frac{ x^{4}-x^{3}+x^{2}-x+2 }{ x-2 } = \color{blue}{x^{3}+x^{2}+3x+5} ~+~ \frac{ \color{red}{ 12 } }{ x-2 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&-1&1&-1&2\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}2&\color{orangered}{ 1 }&-1&1&-1&2\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&-1&1&-1&2\\& & \color{blue}{2} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 2 } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrrr}2&1&\color{orangered}{ -1 }&1&-1&2\\& & \color{orangered}{2} & & & \\ \hline &1&\color{orangered}{1}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 1 } = \color{blue}{ 2 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&-1&1&-1&2\\& & 2& \color{blue}{2} & & \\ \hline &1&\color{blue}{1}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 2 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrrr}2&1&-1&\color{orangered}{ 1 }&-1&2\\& & 2& \color{orangered}{2} & & \\ \hline &1&1&\color{orangered}{3}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 3 } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&-1&1&-1&2\\& & 2& 2& \color{blue}{6} & \\ \hline &1&1&\color{blue}{3}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 6 } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrrr}2&1&-1&1&\color{orangered}{ -1 }&2\\& & 2& 2& \color{orangered}{6} & \\ \hline &1&1&3&\color{orangered}{5}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 5 } = \color{blue}{ 10 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&-1&1&-1&2\\& & 2& 2& 6& \color{blue}{10} \\ \hline &1&1&3&\color{blue}{5}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 10 } = \color{orangered}{ 12 } $
$$ \begin{array}{c|rrrrr}2&1&-1&1&-1&\color{orangered}{ 2 }\\& & 2& 2& 6& \color{orangered}{10} \\ \hline &\color{blue}{1}&\color{blue}{1}&\color{blue}{3}&\color{blue}{5}&\color{orangered}{12} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}+x^{2}+3x+5 } $ with a remainder of $ \color{red}{ 12 } $.