Divide using synthetic division $$ ( x^{4}-8x^{3}+7x^{2}+2x-6 ) \div ( x-7 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}7&1&-8&7&2&-6\\& & 7& -7& 0& \color{black}{14} \\ \hline &\color{blue}{1}&\color{blue}{-1}&\color{blue}{0}&\color{blue}{2}&\color{orangered}{8} \end{array} $$The solution is:
$$ \frac{ x^{4}-8x^{3}+7x^{2}+2x-6 }{ x-7 } = \color{blue}{x^{3}-x^{2}+2} ~+~ \frac{ \color{red}{ 8 } }{ x-7 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -7 = 0 $ ( $ x = \color{blue}{ 7 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{7}&1&-8&7&2&-6\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}7&\color{orangered}{ 1 }&-8&7&2&-6\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 7 } \cdot \color{blue}{ 1 } = \color{blue}{ 7 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{7}&1&-8&7&2&-6\\& & \color{blue}{7} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -8 } + \color{orangered}{ 7 } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrrr}7&1&\color{orangered}{ -8 }&7&2&-6\\& & \color{orangered}{7} & & & \\ \hline &1&\color{orangered}{-1}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 7 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ -7 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{7}&1&-8&7&2&-6\\& & 7& \color{blue}{-7} & & \\ \hline &1&\color{blue}{-1}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ \left( -7 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}7&1&-8&\color{orangered}{ 7 }&2&-6\\& & 7& \color{orangered}{-7} & & \\ \hline &1&-1&\color{orangered}{0}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 7 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{7}&1&-8&7&2&-6\\& & 7& -7& \color{blue}{0} & \\ \hline &1&-1&\color{blue}{0}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 0 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrr}7&1&-8&7&\color{orangered}{ 2 }&-6\\& & 7& -7& \color{orangered}{0} & \\ \hline &1&-1&0&\color{orangered}{2}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 7 } \cdot \color{blue}{ 2 } = \color{blue}{ 14 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{7}&1&-8&7&2&-6\\& & 7& -7& 0& \color{blue}{14} \\ \hline &1&-1&0&\color{blue}{2}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ 14 } = \color{orangered}{ 8 } $
$$ \begin{array}{c|rrrrr}7&1&-8&7&2&\color{orangered}{ -6 }\\& & 7& -7& 0& \color{orangered}{14} \\ \hline &\color{blue}{1}&\color{blue}{-1}&\color{blue}{0}&\color{blue}{2}&\color{orangered}{8} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}-x^{2}+2 } $ with a remainder of $ \color{red}{ 8 } $.