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