Divide using synthetic division $$ ( 7x^{2}+6x+3 ) \div ( x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrr}-3&7&6&3\\& & -21& \color{black}{45} \\ \hline &\color{blue}{7}&\color{blue}{-15}&\color{orangered}{48} \end{array} $$The solution is:
$$ \frac{ 7x^{2}+6x+3 }{ x+3 } = \color{blue}{7x-15} ~+~ \frac{ \color{red}{ 48 } }{ x+3 } $$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.
$$ \begin{array}{c|rrr}\color{blue}{-3}&7&6&3\\& & & \\ \hline &&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrr}-3&\color{orangered}{ 7 }&6&3\\& & & \\ \hline &\color{orangered}{7}&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 7 } = \color{blue}{ -21 } $.
$$ \begin{array}{c|rrr}\color{blue}{-3}&7&6&3\\& & \color{blue}{-21} & \\ \hline &\color{blue}{7}&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ \left( -21 \right) } = \color{orangered}{ -15 } $
$$ \begin{array}{c|rrr}-3&7&\color{orangered}{ 6 }&3\\& & \color{orangered}{-21} & \\ \hline &7&\color{orangered}{-15}& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -15 \right) } = \color{blue}{ 45 } $.
$$ \begin{array}{c|rrr}\color{blue}{-3}&7&6&3\\& & -21& \color{blue}{45} \\ \hline &7&\color{blue}{-15}& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 45 } = \color{orangered}{ 48 } $
$$ \begin{array}{c|rrr}-3&7&6&\color{orangered}{ 3 }\\& & -21& \color{orangered}{45} \\ \hline &\color{blue}{7}&\color{blue}{-15}&\color{orangered}{48} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 7x-15 } $ with a remainder of $ \color{red}{ 48 } $.