Divide using synthetic division $$ ( x^{5}-3x^{4}+5x^{3}-18x^{2}+6x-7 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}3&1&-3&5&-18&6&-7\\& & 3& 0& 15& -9& \color{black}{-9} \\ \hline &\color{blue}{1}&\color{blue}{0}&\color{blue}{5}&\color{blue}{-3}&\color{blue}{-3}&\color{orangered}{-16} \end{array} $$The solution is:
$$ \frac{ x^{5}-3x^{4}+5x^{3}-18x^{2}+6x-7 }{ x-3 } = \color{blue}{x^{4}+5x^{2}-3x-3} \color{red}{~-~} \frac{ \color{red}{ 16 } }{ x-3 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -3 = 0 $ ( $ x = \color{blue}{ 3 } $ ) at the left.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&1&-3&5&-18&6&-7\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}3&\color{orangered}{ 1 }&-3&5&-18&6&-7\\& & & & & & \\ \hline &\color{orangered}{1}&&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 1 } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&1&-3&5&-18&6&-7\\& & \color{blue}{3} & & & & \\ \hline &\color{blue}{1}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ 3 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrrr}3&1&\color{orangered}{ -3 }&5&-18&6&-7\\& & \color{orangered}{3} & & & & \\ \hline &1&\color{orangered}{0}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&1&-3&5&-18&6&-7\\& & 3& \color{blue}{0} & & & \\ \hline &1&\color{blue}{0}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ 0 } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrrrr}3&1&-3&\color{orangered}{ 5 }&-18&6&-7\\& & 3& \color{orangered}{0} & & & \\ \hline &1&0&\color{orangered}{5}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 5 } = \color{blue}{ 15 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&1&-3&5&-18&6&-7\\& & 3& 0& \color{blue}{15} & & \\ \hline &1&0&\color{blue}{5}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -18 } + \color{orangered}{ 15 } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrrrr}3&1&-3&5&\color{orangered}{ -18 }&6&-7\\& & 3& 0& \color{orangered}{15} & & \\ \hline &1&0&5&\color{orangered}{-3}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ -9 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&1&-3&5&-18&6&-7\\& & 3& 0& 15& \color{blue}{-9} & \\ \hline &1&0&5&\color{blue}{-3}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ \left( -9 \right) } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrrrr}3&1&-3&5&-18&\color{orangered}{ 6 }&-7\\& & 3& 0& 15& \color{orangered}{-9} & \\ \hline &1&0&5&-3&\color{orangered}{-3}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ -9 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&1&-3&5&-18&6&-7\\& & 3& 0& 15& -9& \color{blue}{-9} \\ \hline &1&0&5&-3&\color{blue}{-3}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ \left( -9 \right) } = \color{orangered}{ -16 } $
$$ \begin{array}{c|rrrrrr}3&1&-3&5&-18&6&\color{orangered}{ -7 }\\& & 3& 0& 15& -9& \color{orangered}{-9} \\ \hline &\color{blue}{1}&\color{blue}{0}&\color{blue}{5}&\color{blue}{-3}&\color{blue}{-3}&\color{orangered}{-16} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{4}+5x^{2}-3x-3 } $ with a remainder of $ \color{red}{ -16 } $.