Divide using synthetic division $$ ( x^{4}+8x^{3}-16x^{2}+14x-1 ) \div ( x-1 ) $$
The synthetic division table is:
$$ \begin{array}{c|rrrrr}1&1&8&-16&14&-1\\& & 1& 9& -7& \color{black}{7} \\ \hline &\color{blue}{1}&\color{blue}{9}&\color{blue}{-7}&\color{blue}{7}&\color{orangered}{6} \end{array} $$The solution is:
$$ \dfrac{ x^{4}+8x^{3}-16x^{2}+14x-1 }{ x-1 } = \color{blue}{x^{3}+9x^{2}-7x+7} ~+~ \dfrac{ \color{red}{ 6 } }{ x-1 } $$Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -1 = 0 $ ( $ x = \color{blue}{ 1 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&1&8&-16&14&-1\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}1&\color{orangered}{ 1 }&8&-16&14&-1\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 1 } = \color{blue}{ 1 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&1&8&-16&14&-1\\& & \color{blue}{1} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 8 } + \color{orangered}{ 1 } = \color{orangered}{ 9 } $
$$ \begin{array}{c|rrrrr}1&1&\color{orangered}{ 8 }&-16&14&-1\\& & \color{orangered}{1} & & & \\ \hline &1&\color{orangered}{9}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 9 } = \color{blue}{ 9 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&1&8&-16&14&-1\\& & 1& \color{blue}{9} & & \\ \hline &1&\color{blue}{9}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -16 } + \color{orangered}{ 9 } = \color{orangered}{ -7 } $
$$ \begin{array}{c|rrrrr}1&1&8&\color{orangered}{ -16 }&14&-1\\& & 1& \color{orangered}{9} & & \\ \hline &1&9&\color{orangered}{-7}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -7 \right) } = \color{blue}{ -7 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&1&8&-16&14&-1\\& & 1& 9& \color{blue}{-7} & \\ \hline &1&9&\color{blue}{-7}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 14 } + \color{orangered}{ \left( -7 \right) } = \color{orangered}{ 7 } $
$$ \begin{array}{c|rrrrr}1&1&8&-16&\color{orangered}{ 14 }&-1\\& & 1& 9& \color{orangered}{-7} & \\ \hline &1&9&-7&\color{orangered}{7}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 7 } = \color{blue}{ 7 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&1&8&-16&14&-1\\& & 1& 9& -7& \color{blue}{7} \\ \hline &1&9&-7&\color{blue}{7}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 7 } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrrr}1&1&8&-16&14&\color{orangered}{ -1 }\\& & 1& 9& -7& \color{orangered}{7} \\ \hline &\color{blue}{1}&\color{blue}{9}&\color{blue}{-7}&\color{blue}{7}&\color{orangered}{6} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}+9x^{2}-7x+7 } $ with a remainder of $ \color{red}{ 6 } $.