Divide using synthetic division $$ ( x^{5}-7x^{4}+12x^{3}+7x^{2}-25x+23 ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}2&1&-7&12&7&-25&23\\& & 2& -10& 4& 22& \color{black}{-6} \\ \hline &\color{blue}{1}&\color{blue}{-5}&\color{blue}{2}&\color{blue}{11}&\color{blue}{-3}&\color{orangered}{17} \end{array} $$The solution is:
$$ \frac{ x^{5}-7x^{4}+12x^{3}+7x^{2}-25x+23 }{ x-2 } = \color{blue}{x^{4}-5x^{3}+2x^{2}+11x-3} ~+~ \frac{ \color{red}{ 17 } }{ x-2 } $$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.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&1&-7&12&7&-25&23\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}2&\color{orangered}{ 1 }&-7&12&7&-25&23\\& & & & & & \\ \hline &\color{orangered}{1}&&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&1&-7&12&7&-25&23\\& & \color{blue}{2} & & & & \\ \hline &\color{blue}{1}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 2 } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrrrr}2&1&\color{orangered}{ -7 }&12&7&-25&23\\& & \color{orangered}{2} & & & & \\ \hline &1&\color{orangered}{-5}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ -10 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&1&-7&12&7&-25&23\\& & 2& \color{blue}{-10} & & & \\ \hline &1&\color{blue}{-5}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ \left( -10 \right) } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrrr}2&1&-7&\color{orangered}{ 12 }&7&-25&23\\& & 2& \color{orangered}{-10} & & & \\ \hline &1&-5&\color{orangered}{2}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 2 } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&1&-7&12&7&-25&23\\& & 2& -10& \color{blue}{4} & & \\ \hline &1&-5&\color{blue}{2}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ 4 } = \color{orangered}{ 11 } $
$$ \begin{array}{c|rrrrrr}2&1&-7&12&\color{orangered}{ 7 }&-25&23\\& & 2& -10& \color{orangered}{4} & & \\ \hline &1&-5&2&\color{orangered}{11}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 11 } = \color{blue}{ 22 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&1&-7&12&7&-25&23\\& & 2& -10& 4& \color{blue}{22} & \\ \hline &1&-5&2&\color{blue}{11}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -25 } + \color{orangered}{ 22 } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrrrr}2&1&-7&12&7&\color{orangered}{ -25 }&23\\& & 2& -10& 4& \color{orangered}{22} & \\ \hline &1&-5&2&11&\color{orangered}{-3}& \end{array} $$Step 10 : 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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&1&-7&12&7&-25&23\\& & 2& -10& 4& 22& \color{blue}{-6} \\ \hline &1&-5&2&11&\color{blue}{-3}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ 23 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ 17 } $
$$ \begin{array}{c|rrrrrr}2&1&-7&12&7&-25&\color{orangered}{ 23 }\\& & 2& -10& 4& 22& \color{orangered}{-6} \\ \hline &\color{blue}{1}&\color{blue}{-5}&\color{blue}{2}&\color{blue}{11}&\color{blue}{-3}&\color{orangered}{17} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{4}-5x^{3}+2x^{2}+11x-3 } $ with a remainder of $ \color{red}{ 17 } $.