Divide using synthetic division $$ ( 6x^{3}-25x^{2}-31x+20 ) \div ( 3x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}2&6&-25&-31&20\\& & 12& -26& \color{black}{-114} \\ \hline &\color{blue}{6}&\color{blue}{-13}&\color{blue}{-57}&\color{orangered}{-94} \end{array} $$The solution is:
$$ \frac{ 6x^{3}-25x^{2}-31x+20 }{ x-2 } = \color{blue}{6x^{2}-13x-57} \color{red}{~-~} \frac{ \color{red}{ 94 } }{ x-2 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{2}&6&-25&-31&20\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}2&\color{orangered}{ 6 }&-25&-31&20\\& & & & \\ \hline &\color{orangered}{6}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&6&-25&-31&20\\& & \color{blue}{12} & & \\ \hline &\color{blue}{6}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -25 } + \color{orangered}{ 12 } = \color{orangered}{ -13 } $
$$ \begin{array}{c|rrrr}2&6&\color{orangered}{ -25 }&-31&20\\& & \color{orangered}{12} & & \\ \hline &6&\color{orangered}{-13}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -13 \right) } = \color{blue}{ -26 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&6&-25&-31&20\\& & 12& \color{blue}{-26} & \\ \hline &6&\color{blue}{-13}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -31 } + \color{orangered}{ \left( -26 \right) } = \color{orangered}{ -57 } $
$$ \begin{array}{c|rrrr}2&6&-25&\color{orangered}{ -31 }&20\\& & 12& \color{orangered}{-26} & \\ \hline &6&-13&\color{orangered}{-57}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -57 \right) } = \color{blue}{ -114 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&6&-25&-31&20\\& & 12& -26& \color{blue}{-114} \\ \hline &6&-13&\color{blue}{-57}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 20 } + \color{orangered}{ \left( -114 \right) } = \color{orangered}{ -94 } $
$$ \begin{array}{c|rrrr}2&6&-25&-31&\color{orangered}{ 20 }\\& & 12& -26& \color{orangered}{-114} \\ \hline &\color{blue}{6}&\color{blue}{-13}&\color{blue}{-57}&\color{orangered}{-94} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 6x^{2}-13x-57 } $ with a remainder of $ \color{red}{ -94 } $.