Divide using synthetic division $$ ( x^{4}+x^{3}+2x^{2}+27x+7 ) \div ( x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-3&1&1&2&27&7\\& & -3& 6& -24& \color{black}{-9} \\ \hline &\color{blue}{1}&\color{blue}{-2}&\color{blue}{8}&\color{blue}{3}&\color{orangered}{-2} \end{array} $$The solution is:
$$ \frac{ x^{4}+x^{3}+2x^{2}+27x+7 }{ x+3 } = \color{blue}{x^{3}-2x^{2}+8x+3} \color{red}{~-~} \frac{ \color{red}{ 2 } }{ x+3 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&1&1&2&27&7\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-3&\color{orangered}{ 1 }&1&2&27&7\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&1&1&2&27&7\\& & \color{blue}{-3} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrrr}-3&1&\color{orangered}{ 1 }&2&27&7\\& & \color{orangered}{-3} & & & \\ \hline &1&\color{orangered}{-2}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&1&1&2&27&7\\& & -3& \color{blue}{6} & & \\ \hline &1&\color{blue}{-2}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 6 } = \color{orangered}{ 8 } $
$$ \begin{array}{c|rrrrr}-3&1&1&\color{orangered}{ 2 }&27&7\\& & -3& \color{orangered}{6} & & \\ \hline &1&-2&\color{orangered}{8}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 8 } = \color{blue}{ -24 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&1&1&2&27&7\\& & -3& 6& \color{blue}{-24} & \\ \hline &1&-2&\color{blue}{8}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 27 } + \color{orangered}{ \left( -24 \right) } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrrr}-3&1&1&2&\color{orangered}{ 27 }&7\\& & -3& 6& \color{orangered}{-24} & \\ \hline &1&-2&8&\color{orangered}{3}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 3 } = \color{blue}{ -9 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&1&1&2&27&7\\& & -3& 6& -24& \color{blue}{-9} \\ \hline &1&-2&8&\color{blue}{3}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ \left( -9 \right) } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrrr}-3&1&1&2&27&\color{orangered}{ 7 }\\& & -3& 6& -24& \color{orangered}{-9} \\ \hline &\color{blue}{1}&\color{blue}{-2}&\color{blue}{8}&\color{blue}{3}&\color{orangered}{-2} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}-2x^{2}+8x+3 } $ with a remainder of $ \color{red}{ -2 } $.