Divide using synthetic division $$ ( 2x^{3}+9x^{2}+7x-6 ) \div ( x-1 ) $$
The synthetic division table is:
$$ \begin{array}{c|rrrr}1&2&9&7&-6\\& & 2& 11& \color{black}{18} \\ \hline &\color{blue}{2}&\color{blue}{11}&\color{blue}{18}&\color{orangered}{12} \end{array} $$The solution is:
$$ \dfrac{ 2x^{3}+9x^{2}+7x-6 }{ x-1 } = \color{blue}{2x^{2}+11x+18} ~+~ \dfrac{ \color{red}{ 12 } }{ x-1 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{1}&2&9&7&-6\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}1&\color{orangered}{ 2 }&9&7&-6\\& & & & \\ \hline &\color{orangered}{2}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&2&9&7&-6\\& & \color{blue}{2} & & \\ \hline &\color{blue}{2}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 9 } + \color{orangered}{ 2 } = \color{orangered}{ 11 } $
$$ \begin{array}{c|rrrr}1&2&\color{orangered}{ 9 }&7&-6\\& & \color{orangered}{2} & & \\ \hline &2&\color{orangered}{11}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 11 } = \color{blue}{ 11 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&2&9&7&-6\\& & 2& \color{blue}{11} & \\ \hline &2&\color{blue}{11}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ 11 } = \color{orangered}{ 18 } $
$$ \begin{array}{c|rrrr}1&2&9&\color{orangered}{ 7 }&-6\\& & 2& \color{orangered}{11} & \\ \hline &2&11&\color{orangered}{18}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 18 } = \color{blue}{ 18 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&2&9&7&-6\\& & 2& 11& \color{blue}{18} \\ \hline &2&11&\color{blue}{18}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ 18 } = \color{orangered}{ 12 } $
$$ \begin{array}{c|rrrr}1&2&9&7&\color{orangered}{ -6 }\\& & 2& 11& \color{orangered}{18} \\ \hline &\color{blue}{2}&\color{blue}{11}&\color{blue}{18}&\color{orangered}{12} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{2}+11x+18 } $ with a remainder of $ \color{red}{ 12 } $.