Divide using synthetic division $$ ( 2x^{3}+x^{2}-13x+6 ) \div ( x-2 ) $$
The synthetic division table is:
$$ \begin{array}{c|rrrr}2&2&1&-13&6\\& & 4& 10& \color{black}{-6} \\ \hline &\color{blue}{2}&\color{blue}{5}&\color{blue}{-3}&\color{orangered}{0} \end{array} $$The solution is:
$$ \dfrac{ 2x^{3}+x^{2}-13x+6 }{ x-2 } = \color{blue}{2x^{2}+5x-3} $$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.
$$ \begin{array}{c|rrrr}\color{blue}{2}&2&1&-13&6\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}2&\color{orangered}{ 2 }&1&-13&6\\& & & & \\ \hline &\color{orangered}{2}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 2 } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&2&1&-13&6\\& & \color{blue}{4} & & \\ \hline &\color{blue}{2}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 4 } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrr}2&2&\color{orangered}{ 1 }&-13&6\\& & \color{orangered}{4} & & \\ \hline &2&\color{orangered}{5}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 5 } = \color{blue}{ 10 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&2&1&-13&6\\& & 4& \color{blue}{10} & \\ \hline &2&\color{blue}{5}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -13 } + \color{orangered}{ 10 } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrr}2&2&1&\color{orangered}{ -13 }&6\\& & 4& \color{orangered}{10} & \\ \hline &2&5&\color{orangered}{-3}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ -6 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&2&1&-13&6\\& & 4& 10& \color{blue}{-6} \\ \hline &2&5&\color{blue}{-3}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}2&2&1&-13&\color{orangered}{ 6 }\\& & 4& 10& \color{orangered}{-6} \\ \hline &\color{blue}{2}&\color{blue}{5}&\color{blue}{-3}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{2}+5x-3 } $ with a remainder of $ \color{red}{ 0 } $.