Divide using synthetic division $$ ( 6x^{4}+21x^{3}+9x^{2}+8 ) \div ( x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-3&6&21&9&0&8\\& & -18& -9& 0& \color{black}{0} \\ \hline &\color{blue}{6}&\color{blue}{3}&\color{blue}{0}&\color{blue}{0}&\color{orangered}{8} \end{array} $$The solution is:
$$ \frac{ 6x^{4}+21x^{3}+9x^{2}+8 }{ x+3 } = \color{blue}{6x^{3}+3x^{2}} ~+~ \frac{ \color{red}{ 8 } }{ x+3 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&6&21&9&0&8\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-3&\color{orangered}{ 6 }&21&9&0&8\\& & & & & \\ \hline &\color{orangered}{6}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 6 } = \color{blue}{ -18 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&6&21&9&0&8\\& & \color{blue}{-18} & & & \\ \hline &\color{blue}{6}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 21 } + \color{orangered}{ \left( -18 \right) } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrrr}-3&6&\color{orangered}{ 21 }&9&0&8\\& & \color{orangered}{-18} & & & \\ \hline &6&\color{orangered}{3}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 3 } = \color{blue}{ -9 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&6&21&9&0&8\\& & -18& \color{blue}{-9} & & \\ \hline &6&\color{blue}{3}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 9 } + \color{orangered}{ \left( -9 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}-3&6&21&\color{orangered}{ 9 }&0&8\\& & -18& \color{orangered}{-9} & & \\ \hline &6&3&\color{orangered}{0}&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&6&21&9&0&8\\& & -18& -9& \color{blue}{0} & \\ \hline &6&3&\color{blue}{0}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 0 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}-3&6&21&9&\color{orangered}{ 0 }&8\\& & -18& -9& \color{orangered}{0} & \\ \hline &6&3&0&\color{orangered}{0}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&6&21&9&0&8\\& & -18& -9& 0& \color{blue}{0} \\ \hline &6&3&0&\color{blue}{0}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 8 } + \color{orangered}{ 0 } = \color{orangered}{ 8 } $
$$ \begin{array}{c|rrrrr}-3&6&21&9&0&\color{orangered}{ 8 }\\& & -18& -9& 0& \color{orangered}{0} \\ \hline &\color{blue}{6}&\color{blue}{3}&\color{blue}{0}&\color{blue}{0}&\color{orangered}{8} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 6x^{3}+3x^{2} } $ with a remainder of $ \color{red}{ 8 } $.