Divide using synthetic division $$ ( 9x^{3}-6x^{2}+3x-4 ) \div ( x-0.6 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}\frac{ 3 }{ 5 }&9&-6&3&-4\\& & \frac{ 27 }{ 5 }& -\frac{ 9 }{ 25 }& \color{black}{\frac{ 198 }{ 125 }} \\ \hline &\color{blue}{9}&\color{blue}{-\frac{ 3 }{ 5 }}&\color{blue}{\frac{ 66 }{ 25 }}&\color{orangered}{-\frac{ 302 }{ 125 }} \end{array} $$The solution is:
$$ \frac{ 9x^{3}-6x^{2}+3x-4 }{ x-\frac{ 3 }{ 5 } } = \color{blue}{9x^{2}-\frac{ 3 }{ 5 }x+\frac{ 66 }{ 25 }} \color{red}{~-~} \frac{ \color{red}{ \frac{ 302 }{ 125 } } }{ x-\frac{ 3 }{ 5 } } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -\frac{ 3 }{ 5 } = 0 $ ( $ x = \color{blue}{ \frac{ 3 }{ 5 } } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{\frac{ 3 }{ 5 }}&9&-6&3&-4\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}\frac{ 3 }{ 5 }&\color{orangered}{ 9 }&-6&3&-4\\& & & & \\ \hline &\color{orangered}{9}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ \frac{ 3 }{ 5 } } \cdot \color{blue}{ 9 } = \color{blue}{ \frac{ 27 }{ 5 } } $.
$$ \begin{array}{c|rrrr}\color{blue}{\frac{ 3 }{ 5 }}&9&-6&3&-4\\& & \color{blue}{\frac{ 27 }{ 5 }} & & \\ \hline &\color{blue}{9}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ \frac{ 27 }{ 5 } } = \color{orangered}{ -\frac{ 3 }{ 5 } } $
$$ \begin{array}{c|rrrr}\frac{ 3 }{ 5 }&9&\color{orangered}{ -6 }&3&-4\\& & \color{orangered}{\frac{ 27 }{ 5 }} & & \\ \hline &9&\color{orangered}{-\frac{ 3 }{ 5 }}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ \frac{ 3 }{ 5 } } \cdot \color{blue}{ \left( -\frac{ 3 }{ 5 } \right) } = \color{blue}{ -\frac{ 9 }{ 25 } } $.
$$ \begin{array}{c|rrrr}\color{blue}{\frac{ 3 }{ 5 }}&9&-6&3&-4\\& & \frac{ 27 }{ 5 }& \color{blue}{-\frac{ 9 }{ 25 }} & \\ \hline &9&\color{blue}{-\frac{ 3 }{ 5 }}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ \left( -\frac{ 9 }{ 25 } \right) } = \color{orangered}{ \frac{ 66 }{ 25 } } $
$$ \begin{array}{c|rrrr}\frac{ 3 }{ 5 }&9&-6&\color{orangered}{ 3 }&-4\\& & \frac{ 27 }{ 5 }& \color{orangered}{-\frac{ 9 }{ 25 }} & \\ \hline &9&-\frac{ 3 }{ 5 }&\color{orangered}{\frac{ 66 }{ 25 }}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ \frac{ 3 }{ 5 } } \cdot \color{blue}{ \frac{ 66 }{ 25 } } = \color{blue}{ \frac{ 198 }{ 125 } } $.
$$ \begin{array}{c|rrrr}\color{blue}{\frac{ 3 }{ 5 }}&9&-6&3&-4\\& & \frac{ 27 }{ 5 }& -\frac{ 9 }{ 25 }& \color{blue}{\frac{ 198 }{ 125 }} \\ \hline &9&-\frac{ 3 }{ 5 }&\color{blue}{\frac{ 66 }{ 25 }}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ \frac{ 198 }{ 125 } } = \color{orangered}{ -\frac{ 302 }{ 125 } } $
$$ \begin{array}{c|rrrr}\frac{ 3 }{ 5 }&9&-6&3&\color{orangered}{ -4 }\\& & \frac{ 27 }{ 5 }& -\frac{ 9 }{ 25 }& \color{orangered}{\frac{ 198 }{ 125 }} \\ \hline &\color{blue}{9}&\color{blue}{-\frac{ 3 }{ 5 }}&\color{blue}{\frac{ 66 }{ 25 }}&\color{orangered}{-\frac{ 302 }{ 125 }} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 9x^{2}-\frac{ 3 }{ 5 }x+\frac{ 66 }{ 25 } } $ with a remainder of $ \color{red}{ -\frac{ 302 }{ 125 } } $.