Divide using synthetic division $$ ( 2x^{4}+7x^{3}-2x^{2}-13x+6 ) \div ( x-1 ) $$
The synthetic division table is:
$$ \begin{array}{c|rrrrr}1&2&7&-2&-13&6\\& & 2& 9& 7& \color{black}{-6} \\ \hline &\color{blue}{2}&\color{blue}{9}&\color{blue}{7}&\color{blue}{-6}&\color{orangered}{0} \end{array} $$The solution is:
$$ \dfrac{ 2x^{4}+7x^{3}-2x^{2}-13x+6 }{ x-1 } = \color{blue}{2x^{3}+9x^{2}+7x-6} $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&2&7&-2&-13&6\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}1&\color{orangered}{ 2 }&7&-2&-13&6\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 2 } = \color{blue}{ 2 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&2&7&-2&-13&6\\& & \color{blue}{2} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ 2 } = \color{orangered}{ 9 } $
$$ \begin{array}{c|rrrrr}1&2&\color{orangered}{ 7 }&-2&-13&6\\& & \color{orangered}{2} & & & \\ \hline &2&\color{orangered}{9}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&2&7&-2&-13&6\\& & 2& \color{blue}{9} & & \\ \hline &2&\color{blue}{9}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ 9 } = \color{orangered}{ 7 } $
$$ \begin{array}{c|rrrrr}1&2&7&\color{orangered}{ -2 }&-13&6\\& & 2& \color{orangered}{9} & & \\ \hline &2&9&\color{orangered}{7}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 7 } = \color{blue}{ 7 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&2&7&-2&-13&6\\& & 2& 9& \color{blue}{7} & \\ \hline &2&9&\color{blue}{7}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -13 } + \color{orangered}{ 7 } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrrr}1&2&7&-2&\color{orangered}{ -13 }&6\\& & 2& 9& \color{orangered}{7} & \\ \hline &2&9&7&\color{orangered}{-6}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -6 \right) } = \color{blue}{ -6 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&2&7&-2&-13&6\\& & 2& 9& 7& \color{blue}{-6} \\ \hline &2&9&7&\color{blue}{-6}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}1&2&7&-2&-13&\color{orangered}{ 6 }\\& & 2& 9& 7& \color{orangered}{-6} \\ \hline &\color{blue}{2}&\color{blue}{9}&\color{blue}{7}&\color{blue}{-6}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{3}+9x^{2}+7x-6 } $ with a remainder of $ \color{red}{ 0 } $.