Divide using synthetic division $$ ( x^{5}-7x^{4}+10x^{3}+9x-23 ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}2&1&-7&10&0&9&-23\\& & 2& -10& 0& 0& \color{black}{18} \\ \hline &\color{blue}{1}&\color{blue}{-5}&\color{blue}{0}&\color{blue}{0}&\color{blue}{9}&\color{orangered}{-5} \end{array} $$The solution is:
$$ \frac{ x^{5}-7x^{4}+10x^{3}+9x-23 }{ x-2 } = \color{blue}{x^{4}-5x^{3}+9} \color{red}{~-~} \frac{ \color{red}{ 5 } }{ x-2 } $$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.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&1&-7&10&0&9&-23\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}2&\color{orangered}{ 1 }&-7&10&0&9&-23\\& & & & & & \\ \hline &\color{orangered}{1}&&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&1&-7&10&0&9&-23\\& & \color{blue}{2} & & & & \\ \hline &\color{blue}{1}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 2 } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrrrr}2&1&\color{orangered}{ -7 }&10&0&9&-23\\& & \color{orangered}{2} & & & & \\ \hline &1&\color{orangered}{-5}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&1&-7&10&0&9&-23\\& & 2& \color{blue}{-10} & & & \\ \hline &1&\color{blue}{-5}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 10 } + \color{orangered}{ \left( -10 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrrr}2&1&-7&\color{orangered}{ 10 }&0&9&-23\\& & 2& \color{orangered}{-10} & & & \\ \hline &1&-5&\color{orangered}{0}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&1&-7&10&0&9&-23\\& & 2& -10& \color{blue}{0} & & \\ \hline &1&-5&\color{blue}{0}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 0 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrrr}2&1&-7&10&\color{orangered}{ 0 }&9&-23\\& & 2& -10& \color{orangered}{0} & & \\ \hline &1&-5&0&\color{orangered}{0}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&1&-7&10&0&9&-23\\& & 2& -10& 0& \color{blue}{0} & \\ \hline &1&-5&0&\color{blue}{0}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 9 } + \color{orangered}{ 0 } = \color{orangered}{ 9 } $
$$ \begin{array}{c|rrrrrr}2&1&-7&10&0&\color{orangered}{ 9 }&-23\\& & 2& -10& 0& \color{orangered}{0} & \\ \hline &1&-5&0&0&\color{orangered}{9}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 9 } = \color{blue}{ 18 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&1&-7&10&0&9&-23\\& & 2& -10& 0& 0& \color{blue}{18} \\ \hline &1&-5&0&0&\color{blue}{9}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ -23 } + \color{orangered}{ 18 } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrrrr}2&1&-7&10&0&9&\color{orangered}{ -23 }\\& & 2& -10& 0& 0& \color{orangered}{18} \\ \hline &\color{blue}{1}&\color{blue}{-5}&\color{blue}{0}&\color{blue}{0}&\color{blue}{9}&\color{orangered}{-5} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{4}-5x^{3}+9 } $ with a remainder of $ \color{red}{ -5 } $.