Divide using synthetic division $$ ( x^{5}+9x^{3}-12x^{2}+16x-18 ) \div ( 3x-1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}1&1&0&9&-12&16&-18\\& & 1& 1& 10& -2& \color{black}{14} \\ \hline &\color{blue}{1}&\color{blue}{1}&\color{blue}{10}&\color{blue}{-2}&\color{blue}{14}&\color{orangered}{-4} \end{array} $$The solution is:
$$ \frac{ x^{5}+9x^{3}-12x^{2}+16x-18 }{ x-1 } = \color{blue}{x^{4}+x^{3}+10x^{2}-2x+14} \color{red}{~-~} \frac{ \color{red}{ 4 } }{ x-1 } $$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.
$$ \begin{array}{c|rrrrrr}\color{blue}{1}&1&0&9&-12&16&-18\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}1&\color{orangered}{ 1 }&0&9&-12&16&-18\\& & & & & & \\ \hline &\color{orangered}{1}&&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 1 } = \color{blue}{ 1 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{1}&1&0&9&-12&16&-18\\& & \color{blue}{1} & & & & \\ \hline &\color{blue}{1}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 1 } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrrrr}1&1&\color{orangered}{ 0 }&9&-12&16&-18\\& & \color{orangered}{1} & & & & \\ \hline &1&\color{orangered}{1}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{1}&1&0&9&-12&16&-18\\& & 1& \color{blue}{1} & & & \\ \hline &1&\color{blue}{1}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 9 } + \color{orangered}{ 1 } = \color{orangered}{ 10 } $
$$ \begin{array}{c|rrrrrr}1&1&0&\color{orangered}{ 9 }&-12&16&-18\\& & 1& \color{orangered}{1} & & & \\ \hline &1&1&\color{orangered}{10}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 10 } = \color{blue}{ 10 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{1}&1&0&9&-12&16&-18\\& & 1& 1& \color{blue}{10} & & \\ \hline &1&1&\color{blue}{10}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -12 } + \color{orangered}{ 10 } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrrrr}1&1&0&9&\color{orangered}{ -12 }&16&-18\\& & 1& 1& \color{orangered}{10} & & \\ \hline &1&1&10&\color{orangered}{-2}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ -2 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{1}&1&0&9&-12&16&-18\\& & 1& 1& 10& \color{blue}{-2} & \\ \hline &1&1&10&\color{blue}{-2}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 16 } + \color{orangered}{ \left( -2 \right) } = \color{orangered}{ 14 } $
$$ \begin{array}{c|rrrrrr}1&1&0&9&-12&\color{orangered}{ 16 }&-18\\& & 1& 1& 10& \color{orangered}{-2} & \\ \hline &1&1&10&-2&\color{orangered}{14}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 14 } = \color{blue}{ 14 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{1}&1&0&9&-12&16&-18\\& & 1& 1& 10& -2& \color{blue}{14} \\ \hline &1&1&10&-2&\color{blue}{14}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ -18 } + \color{orangered}{ 14 } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrrrr}1&1&0&9&-12&16&\color{orangered}{ -18 }\\& & 1& 1& 10& -2& \color{orangered}{14} \\ \hline &\color{blue}{1}&\color{blue}{1}&\color{blue}{10}&\color{blue}{-2}&\color{blue}{14}&\color{orangered}{-4} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{4}+x^{3}+10x^{2}-2x+14 } $ with a remainder of $ \color{red}{ -4 } $.