Divide using synthetic division $$ ( 6x^{3}-16x^{2}+17x-6 ) \div ( 3x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}2&6&-16&17&-6\\& & 12& -8& \color{black}{18} \\ \hline &\color{blue}{6}&\color{blue}{-4}&\color{blue}{9}&\color{orangered}{12} \end{array} $$The solution is:
$$ \frac{ 6x^{3}-16x^{2}+17x-6 }{ x-2 } = \color{blue}{6x^{2}-4x+9} ~+~ \frac{ \color{red}{ 12 } }{ x-2 } $$Explanation
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.
$$ \begin{array}{c|rrrr}\color{blue}{2}&6&-16&17&-6\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}2&\color{orangered}{ 6 }&-16&17&-6\\& & & & \\ \hline &\color{orangered}{6}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 6 } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&6&-16&17&-6\\& & \color{blue}{12} & & \\ \hline &\color{blue}{6}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -16 } + \color{orangered}{ 12 } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrr}2&6&\color{orangered}{ -16 }&17&-6\\& & \color{orangered}{12} & & \\ \hline &6&\color{orangered}{-4}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ -8 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&6&-16&17&-6\\& & 12& \color{blue}{-8} & \\ \hline &6&\color{blue}{-4}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 17 } + \color{orangered}{ \left( -8 \right) } = \color{orangered}{ 9 } $
$$ \begin{array}{c|rrrr}2&6&-16&\color{orangered}{ 17 }&-6\\& & 12& \color{orangered}{-8} & \\ \hline &6&-4&\color{orangered}{9}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 9 } = \color{blue}{ 18 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&6&-16&17&-6\\& & 12& -8& \color{blue}{18} \\ \hline &6&-4&\color{blue}{9}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ 18 } = \color{orangered}{ 12 } $
$$ \begin{array}{c|rrrr}2&6&-16&17&\color{orangered}{ -6 }\\& & 12& -8& \color{orangered}{18} \\ \hline &\color{blue}{6}&\color{blue}{-4}&\color{blue}{9}&\color{orangered}{12} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 6x^{2}-4x+9 } $ with a remainder of $ \color{red}{ 12 } $.