Divide using synthetic division $$ ( 6x^{3}-4x^{2}+3x-2 ) \div ( 2x+1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-1&6&-4&3&-2\\& & -6& 10& \color{black}{-13} \\ \hline &\color{blue}{6}&\color{blue}{-10}&\color{blue}{13}&\color{orangered}{-15} \end{array} $$The solution is:
$$ \frac{ 6x^{3}-4x^{2}+3x-2 }{ x+1 } = \color{blue}{6x^{2}-10x+13} \color{red}{~-~} \frac{ \color{red}{ 15 } }{ x+1 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 1 = 0 $ ( $ x = \color{blue}{ -1 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&6&-4&3&-2\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-1&\color{orangered}{ 6 }&-4&3&-2\\& & & & \\ \hline &\color{orangered}{6}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 6 } = \color{blue}{ -6 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&6&-4&3&-2\\& & \color{blue}{-6} & & \\ \hline &\color{blue}{6}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ -10 } $
$$ \begin{array}{c|rrrr}-1&6&\color{orangered}{ -4 }&3&-2\\& & \color{orangered}{-6} & & \\ \hline &6&\color{orangered}{-10}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ \left( -10 \right) } = \color{blue}{ 10 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&6&-4&3&-2\\& & -6& \color{blue}{10} & \\ \hline &6&\color{blue}{-10}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 10 } = \color{orangered}{ 13 } $
$$ \begin{array}{c|rrrr}-1&6&-4&\color{orangered}{ 3 }&-2\\& & -6& \color{orangered}{10} & \\ \hline &6&-10&\color{orangered}{13}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 13 } = \color{blue}{ -13 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&6&-4&3&-2\\& & -6& 10& \color{blue}{-13} \\ \hline &6&-10&\color{blue}{13}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ \left( -13 \right) } = \color{orangered}{ -15 } $
$$ \begin{array}{c|rrrr}-1&6&-4&3&\color{orangered}{ -2 }\\& & -6& 10& \color{orangered}{-13} \\ \hline &\color{blue}{6}&\color{blue}{-10}&\color{blue}{13}&\color{orangered}{-15} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 6x^{2}-10x+13 } $ with a remainder of $ \color{red}{ -15 } $.