Divide using synthetic division $$ ( 6x^{3}+17x-22 ) \div ( 3x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}2&6&0&17&-22\\& & 12& 24& \color{black}{82} \\ \hline &\color{blue}{6}&\color{blue}{12}&\color{blue}{41}&\color{orangered}{60} \end{array} $$The solution is:
$$ \frac{ 6x^{3}+17x-22 }{ x-2 } = \color{blue}{6x^{2}+12x+41} ~+~ \frac{ \color{red}{ 60 } }{ x-2 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{2}&6&0&17&-22\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}2&\color{orangered}{ 6 }&0&17&-22\\& & & & \\ \hline &\color{orangered}{6}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&6&0&17&-22\\& & \color{blue}{12} & & \\ \hline &\color{blue}{6}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 12 } = \color{orangered}{ 12 } $
$$ \begin{array}{c|rrrr}2&6&\color{orangered}{ 0 }&17&-22\\& & \color{orangered}{12} & & \\ \hline &6&\color{orangered}{12}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 12 } = \color{blue}{ 24 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&6&0&17&-22\\& & 12& \color{blue}{24} & \\ \hline &6&\color{blue}{12}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 17 } + \color{orangered}{ 24 } = \color{orangered}{ 41 } $
$$ \begin{array}{c|rrrr}2&6&0&\color{orangered}{ 17 }&-22\\& & 12& \color{orangered}{24} & \\ \hline &6&12&\color{orangered}{41}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 41 } = \color{blue}{ 82 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&6&0&17&-22\\& & 12& 24& \color{blue}{82} \\ \hline &6&12&\color{blue}{41}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -22 } + \color{orangered}{ 82 } = \color{orangered}{ 60 } $
$$ \begin{array}{c|rrrr}2&6&0&17&\color{orangered}{ -22 }\\& & 12& 24& \color{orangered}{82} \\ \hline &\color{blue}{6}&\color{blue}{12}&\color{blue}{41}&\color{orangered}{60} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 6x^{2}+12x+41 } $ with a remainder of $ \color{red}{ 60 } $.