Divide using synthetic division $$ ( 3x^{3}-6x^{2}-8x+20 ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}2&3&-6&-8&20\\& & 6& 0& \color{black}{-16} \\ \hline &\color{blue}{3}&\color{blue}{0}&\color{blue}{-8}&\color{orangered}{4} \end{array} $$The solution is:
$$ \frac{ 3x^{3}-6x^{2}-8x+20 }{ x-2 } = \color{blue}{3x^{2}-8} ~+~ \frac{ \color{red}{ 4 } }{ x-2 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{2}&3&-6&-8&20\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}2&\color{orangered}{ 3 }&-6&-8&20\\& & & & \\ \hline &\color{orangered}{3}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&3&-6&-8&20\\& & \color{blue}{6} & & \\ \hline &\color{blue}{3}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ 6 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}2&3&\color{orangered}{ -6 }&-8&20\\& & \color{orangered}{6} & & \\ \hline &3&\color{orangered}{0}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&3&-6&-8&20\\& & 6& \color{blue}{0} & \\ \hline &3&\color{blue}{0}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -8 } + \color{orangered}{ 0 } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrr}2&3&-6&\color{orangered}{ -8 }&20\\& & 6& \color{orangered}{0} & \\ \hline &3&0&\color{orangered}{-8}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -8 \right) } = \color{blue}{ -16 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&3&-6&-8&20\\& & 6& 0& \color{blue}{-16} \\ \hline &3&0&\color{blue}{-8}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 20 } + \color{orangered}{ \left( -16 \right) } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrr}2&3&-6&-8&\color{orangered}{ 20 }\\& & 6& 0& \color{orangered}{-16} \\ \hline &\color{blue}{3}&\color{blue}{0}&\color{blue}{-8}&\color{orangered}{4} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{2}-8 } $ with a remainder of $ \color{red}{ 4 } $.