Divide using synthetic division $$ ( x^{3}-3x^{2}+81x-243 ) \div ( x+3 ) $$
The synthetic division table is:
$$ \begin{array}{c|rrrr}-3&1&-3&81&-243\\& & -3& 18& \color{black}{-297} \\ \hline &\color{blue}{1}&\color{blue}{-6}&\color{blue}{99}&\color{orangered}{-540} \end{array} $$The solution is:
$$ \dfrac{ x^{3}-3x^{2}+81x-243 }{ x+3 } = \color{blue}{x^{2}-6x+99} \color{red}{~-~} \dfrac{ \color{red}{ 540 } }{ x+3 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&1&-3&81&-243\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-3&\color{orangered}{ 1 }&-3&81&-243\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&1&-3&81&-243\\& & \color{blue}{-3} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrr}-3&1&\color{orangered}{ -3 }&81&-243\\& & \color{orangered}{-3} & & \\ \hline &1&\color{orangered}{-6}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -6 \right) } = \color{blue}{ 18 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&1&-3&81&-243\\& & -3& \color{blue}{18} & \\ \hline &1&\color{blue}{-6}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 81 } + \color{orangered}{ 18 } = \color{orangered}{ 99 } $
$$ \begin{array}{c|rrrr}-3&1&-3&\color{orangered}{ 81 }&-243\\& & -3& \color{orangered}{18} & \\ \hline &1&-6&\color{orangered}{99}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 99 } = \color{blue}{ -297 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&1&-3&81&-243\\& & -3& 18& \color{blue}{-297} \\ \hline &1&-6&\color{blue}{99}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -243 } + \color{orangered}{ \left( -297 \right) } = \color{orangered}{ -540 } $
$$ \begin{array}{c|rrrr}-3&1&-3&81&\color{orangered}{ -243 }\\& & -3& 18& \color{orangered}{-297} \\ \hline &\color{blue}{1}&\color{blue}{-6}&\color{blue}{99}&\color{orangered}{-540} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}-6x+99 } $ with a remainder of $ \color{red}{ -540 } $.