Divide using synthetic division $$ ( 5x^{4}-18x^{3}+27x-7 ) \div ( x-3 ) $$
The synthetic division table is:
$$ \begin{array}{c|rrrrr}3&5&-18&0&27&-7\\& & 15& -9& -27& \color{black}{0} \\ \hline &\color{blue}{5}&\color{blue}{-3}&\color{blue}{-9}&\color{blue}{0}&\color{orangered}{-7} \end{array} $$The solution is:
$$ \dfrac{ 5x^{4}-18x^{3}+27x-7 }{ x-3 } = \color{blue}{5x^{3}-3x^{2}-9x} \color{red}{~-~} \dfrac{ \color{red}{ 7 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&5&-18&0&27&-7\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}3&\color{orangered}{ 5 }&-18&0&27&-7\\& & & & & \\ \hline &\color{orangered}{5}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 5 } = \color{blue}{ 15 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&5&-18&0&27&-7\\& & \color{blue}{15} & & & \\ \hline &\color{blue}{5}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -18 } + \color{orangered}{ 15 } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrrr}3&5&\color{orangered}{ -18 }&0&27&-7\\& & \color{orangered}{15} & & & \\ \hline &5&\color{orangered}{-3}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ -9 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&5&-18&0&27&-7\\& & 15& \color{blue}{-9} & & \\ \hline &5&\color{blue}{-3}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -9 \right) } = \color{orangered}{ -9 } $
$$ \begin{array}{c|rrrrr}3&5&-18&\color{orangered}{ 0 }&27&-7\\& & 15& \color{orangered}{-9} & & \\ \hline &5&-3&\color{orangered}{-9}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -9 \right) } = \color{blue}{ -27 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&5&-18&0&27&-7\\& & 15& -9& \color{blue}{-27} & \\ \hline &5&-3&\color{blue}{-9}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 27 } + \color{orangered}{ \left( -27 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}3&5&-18&0&\color{orangered}{ 27 }&-7\\& & 15& -9& \color{orangered}{-27} & \\ \hline &5&-3&-9&\color{orangered}{0}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&5&-18&0&27&-7\\& & 15& -9& -27& \color{blue}{0} \\ \hline &5&-3&-9&\color{blue}{0}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 0 } = \color{orangered}{ -7 } $
$$ \begin{array}{c|rrrrr}3&5&-18&0&27&\color{orangered}{ -7 }\\& & 15& -9& -27& \color{orangered}{0} \\ \hline &\color{blue}{5}&\color{blue}{-3}&\color{blue}{-9}&\color{blue}{0}&\color{orangered}{-7} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 5x^{3}-3x^{2}-9x } $ with a remainder of $ \color{red}{ -7 } $.