Divide using synthetic division $$ ( 3x^{4}+19x^{3}+30x^{2}+9x+27 ) \div ( x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-3&3&19&30&9&27\\& & -9& -30& 0& \color{black}{-27} \\ \hline &\color{blue}{3}&\color{blue}{10}&\color{blue}{0}&\color{blue}{9}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 3x^{4}+19x^{3}+30x^{2}+9x+27 }{ x+3 } = \color{blue}{3x^{3}+10x^{2}+9} $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&3&19&30&9&27\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-3&\color{orangered}{ 3 }&19&30&9&27\\& & & & & \\ \hline &\color{orangered}{3}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 3 } = \color{blue}{ -9 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&3&19&30&9&27\\& & \color{blue}{-9} & & & \\ \hline &\color{blue}{3}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 19 } + \color{orangered}{ \left( -9 \right) } = \color{orangered}{ 10 } $
$$ \begin{array}{c|rrrrr}-3&3&\color{orangered}{ 19 }&30&9&27\\& & \color{orangered}{-9} & & & \\ \hline &3&\color{orangered}{10}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 10 } = \color{blue}{ -30 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&3&19&30&9&27\\& & -9& \color{blue}{-30} & & \\ \hline &3&\color{blue}{10}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 30 } + \color{orangered}{ \left( -30 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}-3&3&19&\color{orangered}{ 30 }&9&27\\& & -9& \color{orangered}{-30} & & \\ \hline &3&10&\color{orangered}{0}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&3&19&30&9&27\\& & -9& -30& \color{blue}{0} & \\ \hline &3&10&\color{blue}{0}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 9 } + \color{orangered}{ 0 } = \color{orangered}{ 9 } $
$$ \begin{array}{c|rrrrr}-3&3&19&30&\color{orangered}{ 9 }&27\\& & -9& -30& \color{orangered}{0} & \\ \hline &3&10&0&\color{orangered}{9}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 9 } = \color{blue}{ -27 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&3&19&30&9&27\\& & -9& -30& 0& \color{blue}{-27} \\ \hline &3&10&0&\color{blue}{9}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 27 } + \color{orangered}{ \left( -27 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}-3&3&19&30&9&\color{orangered}{ 27 }\\& & -9& -30& 0& \color{orangered}{-27} \\ \hline &\color{blue}{3}&\color{blue}{10}&\color{blue}{0}&\color{blue}{9}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{3}+10x^{2}+9 } $ with a remainder of $ \color{red}{ 0 } $.