Divide using synthetic division $$ ( 9x^{4}-x^{2}-6x+2 ) \div ( 3x-1 ) $$
The synthetic division table is:
$$ \begin{array}{c|rrrrr}1&9&0&-1&-6&2\\& & 9& 9& 8& \color{black}{2} \\ \hline &\color{blue}{9}&\color{blue}{9}&\color{blue}{8}&\color{blue}{2}&\color{orangered}{4} \end{array} $$The solution is:
$$ \dfrac{ 9x^{4}-x^{2}-6x+2 }{ x-1 } = \color{blue}{9x^{3}+9x^{2}+8x+2} ~+~ \dfrac{ \color{red}{ 4 } }{ x-1 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&9&0&-1&-6&2\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}1&\color{orangered}{ 9 }&0&-1&-6&2\\& & & & & \\ \hline &\color{orangered}{9}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 9 } = \color{blue}{ 9 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&9&0&-1&-6&2\\& & \color{blue}{9} & & & \\ \hline &\color{blue}{9}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 9 } = \color{orangered}{ 9 } $
$$ \begin{array}{c|rrrrr}1&9&\color{orangered}{ 0 }&-1&-6&2\\& & \color{orangered}{9} & & & \\ \hline &9&\color{orangered}{9}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 9 } = \color{blue}{ 9 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&9&0&-1&-6&2\\& & 9& \color{blue}{9} & & \\ \hline &9&\color{blue}{9}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 9 } = \color{orangered}{ 8 } $
$$ \begin{array}{c|rrrrr}1&9&0&\color{orangered}{ -1 }&-6&2\\& & 9& \color{orangered}{9} & & \\ \hline &9&9&\color{orangered}{8}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 8 } = \color{blue}{ 8 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&9&0&-1&-6&2\\& & 9& 9& \color{blue}{8} & \\ \hline &9&9&\color{blue}{8}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ 8 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrr}1&9&0&-1&\color{orangered}{ -6 }&2\\& & 9& 9& \color{orangered}{8} & \\ \hline &9&9&8&\color{orangered}{2}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 2 } = \color{blue}{ 2 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&9&0&-1&-6&2\\& & 9& 9& 8& \color{blue}{2} \\ \hline &9&9&8&\color{blue}{2}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 2 } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrrr}1&9&0&-1&-6&\color{orangered}{ 2 }\\& & 9& 9& 8& \color{orangered}{2} \\ \hline &\color{blue}{9}&\color{blue}{9}&\color{blue}{8}&\color{blue}{2}&\color{orangered}{4} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 9x^{3}+9x^{2}+8x+2 } $ with a remainder of $ \color{red}{ 4 } $.