Divide using synthetic division $$ ( 6x^{2}+21x+15 ) \div ( 3x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrr}-3&6&21&15\\& & -18& \color{black}{-9} \\ \hline &\color{blue}{6}&\color{blue}{3}&\color{orangered}{6} \end{array} $$The solution is:
$$ \frac{ 6x^{2}+21x+15 }{ x+3 } = \color{blue}{6x+3} ~+~ \frac{ \color{red}{ 6 } }{ x+3 } $$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.
$$ \begin{array}{c|rrr}\color{blue}{-3}&6&21&15\\& & & \\ \hline &&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrr}-3&\color{orangered}{ 6 }&21&15\\& & & \\ \hline &\color{orangered}{6}&& \end{array} $$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 } $.
$$ \begin{array}{c|rrr}\color{blue}{-3}&6&21&15\\& & \color{blue}{-18} & \\ \hline &\color{blue}{6}&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 21 } + \color{orangered}{ \left( -18 \right) } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrr}-3&6&\color{orangered}{ 21 }&15\\& & \color{orangered}{-18} & \\ \hline &6&\color{orangered}{3}& \end{array} $$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 } $.
$$ \begin{array}{c|rrr}\color{blue}{-3}&6&21&15\\& & -18& \color{blue}{-9} \\ \hline &6&\color{blue}{3}& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 15 } + \color{orangered}{ \left( -9 \right) } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrr}-3&6&21&\color{orangered}{ 15 }\\& & -18& \color{orangered}{-9} \\ \hline &\color{blue}{6}&\color{blue}{3}&\color{orangered}{6} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 6x+3 } $ with a remainder of $ \color{red}{ 6 } $.