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