Divide using synthetic division $$ ( x^{4}-3x^{3}-11x^{2}+6x-24 ) \div ( x+3 ) $$
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-3&1&-3&-11&6&-24\\& & -3& 18& -21& \color{black}{45} \\ \hline &\color{blue}{1}&\color{blue}{-6}&\color{blue}{7}&\color{blue}{-15}&\color{orangered}{21} \end{array} $$The solution is:
$$ \dfrac{ x^{4}-3x^{3}-11x^{2}+6x-24 }{ x+3 } = \color{blue}{x^{3}-6x^{2}+7x-15} ~+~ \dfrac{ \color{red}{ 21 } }{ x+3 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&1&-3&-11&6&-24\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-3&\color{orangered}{ 1 }&-3&-11&6&-24\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&1&-3&-11&6&-24\\& & \color{blue}{-3} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrrr}-3&1&\color{orangered}{ -3 }&-11&6&-24\\& & \color{orangered}{-3} & & & \\ \hline &1&\color{orangered}{-6}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&1&-3&-11&6&-24\\& & -3& \color{blue}{18} & & \\ \hline &1&\color{blue}{-6}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -11 } + \color{orangered}{ 18 } = \color{orangered}{ 7 } $
$$ \begin{array}{c|rrrrr}-3&1&-3&\color{orangered}{ -11 }&6&-24\\& & -3& \color{orangered}{18} & & \\ \hline &1&-6&\color{orangered}{7}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 7 } = \color{blue}{ -21 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&1&-3&-11&6&-24\\& & -3& 18& \color{blue}{-21} & \\ \hline &1&-6&\color{blue}{7}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ \left( -21 \right) } = \color{orangered}{ -15 } $
$$ \begin{array}{c|rrrrr}-3&1&-3&-11&\color{orangered}{ 6 }&-24\\& & -3& 18& \color{orangered}{-21} & \\ \hline &1&-6&7&\color{orangered}{-15}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -15 \right) } = \color{blue}{ 45 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&1&-3&-11&6&-24\\& & -3& 18& -21& \color{blue}{45} \\ \hline &1&-6&7&\color{blue}{-15}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -24 } + \color{orangered}{ 45 } = \color{orangered}{ 21 } $
$$ \begin{array}{c|rrrrr}-3&1&-3&-11&6&\color{orangered}{ -24 }\\& & -3& 18& -21& \color{orangered}{45} \\ \hline &\color{blue}{1}&\color{blue}{-6}&\color{blue}{7}&\color{blue}{-15}&\color{orangered}{21} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}-6x^{2}+7x-15 } $ with a remainder of $ \color{red}{ 21 } $.