Divide using synthetic division $$ ( x^{4}-6x^{3}-28x^{2}+6x+27 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}3&1&-6&-28&6&27\\& & 3& -9& -111& \color{black}{-315} \\ \hline &\color{blue}{1}&\color{blue}{-3}&\color{blue}{-37}&\color{blue}{-105}&\color{orangered}{-288} \end{array} $$The solution is:
$$ \frac{ x^{4}-6x^{3}-28x^{2}+6x+27 }{ x-3 } = \color{blue}{x^{3}-3x^{2}-37x-105} \color{red}{~-~} \frac{ \color{red}{ 288 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&-6&-28&6&27\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}3&\color{orangered}{ 1 }&-6&-28&6&27\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&-6&-28&6&27\\& & \color{blue}{3} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ 3 } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrrr}3&1&\color{orangered}{ -6 }&-28&6&27\\& & \color{orangered}{3} & & & \\ \hline &1&\color{orangered}{-3}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&-6&-28&6&27\\& & 3& \color{blue}{-9} & & \\ \hline &1&\color{blue}{-3}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -28 } + \color{orangered}{ \left( -9 \right) } = \color{orangered}{ -37 } $
$$ \begin{array}{c|rrrrr}3&1&-6&\color{orangered}{ -28 }&6&27\\& & 3& \color{orangered}{-9} & & \\ \hline &1&-3&\color{orangered}{-37}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -37 \right) } = \color{blue}{ -111 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&-6&-28&6&27\\& & 3& -9& \color{blue}{-111} & \\ \hline &1&-3&\color{blue}{-37}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ \left( -111 \right) } = \color{orangered}{ -105 } $
$$ \begin{array}{c|rrrrr}3&1&-6&-28&\color{orangered}{ 6 }&27\\& & 3& -9& \color{orangered}{-111} & \\ \hline &1&-3&-37&\color{orangered}{-105}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -105 \right) } = \color{blue}{ -315 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&-6&-28&6&27\\& & 3& -9& -111& \color{blue}{-315} \\ \hline &1&-3&-37&\color{blue}{-105}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 27 } + \color{orangered}{ \left( -315 \right) } = \color{orangered}{ -288 } $
$$ \begin{array}{c|rrrrr}3&1&-6&-28&6&\color{orangered}{ 27 }\\& & 3& -9& -111& \color{orangered}{-315} \\ \hline &\color{blue}{1}&\color{blue}{-3}&\color{blue}{-37}&\color{blue}{-105}&\color{orangered}{-288} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}-3x^{2}-37x-105 } $ with a remainder of $ \color{red}{ -288 } $.