Divide using synthetic division $$ ( 3x^{3}-2x-8 ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}2&3&0&-2&-8\\& & 6& 12& \color{black}{20} \\ \hline &\color{blue}{3}&\color{blue}{6}&\color{blue}{10}&\color{orangered}{12} \end{array} $$The solution is:
$$ \frac{ 3x^{3}-2x-8 }{ x-2 } = \color{blue}{3x^{2}+6x+10} ~+~ \frac{ \color{red}{ 12 } }{ x-2 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{2}&3&0&-2&-8\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}2&\color{orangered}{ 3 }&0&-2&-8\\& & & & \\ \hline &\color{orangered}{3}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 3 } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&3&0&-2&-8\\& & \color{blue}{6} & & \\ \hline &\color{blue}{3}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 6 } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrr}2&3&\color{orangered}{ 0 }&-2&-8\\& & \color{orangered}{6} & & \\ \hline &3&\color{orangered}{6}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 6 } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&3&0&-2&-8\\& & 6& \color{blue}{12} & \\ \hline &3&\color{blue}{6}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ 12 } = \color{orangered}{ 10 } $
$$ \begin{array}{c|rrrr}2&3&0&\color{orangered}{ -2 }&-8\\& & 6& \color{orangered}{12} & \\ \hline &3&6&\color{orangered}{10}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 10 } = \color{blue}{ 20 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&3&0&-2&-8\\& & 6& 12& \color{blue}{20} \\ \hline &3&6&\color{blue}{10}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -8 } + \color{orangered}{ 20 } = \color{orangered}{ 12 } $
$$ \begin{array}{c|rrrr}2&3&0&-2&\color{orangered}{ -8 }\\& & 6& 12& \color{orangered}{20} \\ \hline &\color{blue}{3}&\color{blue}{6}&\color{blue}{10}&\color{orangered}{12} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{2}+6x+10 } $ with a remainder of $ \color{red}{ 12 } $.