Divide using synthetic division $$ ( 6x^{3}+5x^{2}+4x-1 ) \div ( 3x-1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}1&6&5&4&-1\\& & 6& 11& \color{black}{15} \\ \hline &\color{blue}{6}&\color{blue}{11}&\color{blue}{15}&\color{orangered}{14} \end{array} $$The solution is:
$$ \frac{ 6x^{3}+5x^{2}+4x-1 }{ x-1 } = \color{blue}{6x^{2}+11x+15} ~+~ \frac{ \color{red}{ 14 } }{ x-1 } $$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}{ 6 } = \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}{ 11 } = \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}{ 15 } = \color{orangered}{ 14 } $
$$ \begin{array}{c|rrrr}1&6&5&4&\color{orangered}{ -1 }\\& & 6& 11& \color{orangered}{15} \\ \hline &\color{blue}{6}&\color{blue}{11}&\color{blue}{15}&\color{orangered}{14} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 6x^{2}+11x+15 } $ with a remainder of $ \color{red}{ 14 } $.