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