Divide using synthetic division $$ ( 6x^{3}-5x^{2}+4x-1 ) \div ( 3x+1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-1&6&-5&4&-1\\& & -6& 11& \color{black}{-15} \\ \hline &\color{blue}{6}&\color{blue}{-11}&\color{blue}{15}&\color{orangered}{-16} \end{array} $$The solution is:
$$ \frac{ 6x^{3}-5x^{2}+4x-1 }{ x+1 } = \color{blue}{6x^{2}-11x+15} \color{red}{~-~} \frac{ \color{red}{ 16 } }{ x+1 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&6&-5&4&-1\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-1&\color{orangered}{ 6 }&-5&4&-1\\& & & & \\ \hline &\color{orangered}{6}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&6&-5&4&-1\\& & \color{blue}{-6} & & \\ \hline &\color{blue}{6}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ -11 } $
$$ \begin{array}{c|rrrr}-1&6&\color{orangered}{ -5 }&4&-1\\& & \color{orangered}{-6} & & \\ \hline &6&\color{orangered}{-11}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ \left( -11 \right) } = \color{blue}{ 11 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&6&-5&4&-1\\& & -6& \color{blue}{11} & \\ \hline &6&\color{blue}{-11}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ 11 } = \color{orangered}{ 15 } $
$$ \begin{array}{c|rrrr}-1&6&-5&\color{orangered}{ 4 }&-1\\& & -6& \color{orangered}{11} & \\ \hline &6&-11&\color{orangered}{15}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 15 } = \color{blue}{ -15 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&6&-5&4&-1\\& & -6& 11& \color{blue}{-15} \\ \hline &6&-11&\color{blue}{15}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ \left( -15 \right) } = \color{orangered}{ -16 } $
$$ \begin{array}{c|rrrr}-1&6&-5&4&\color{orangered}{ -1 }\\& & -6& 11& \color{orangered}{-15} \\ \hline &\color{blue}{6}&\color{blue}{-11}&\color{blue}{15}&\color{orangered}{-16} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 6x^{2}-11x+15 } $ with a remainder of $ \color{red}{ -16 } $.