Divide using synthetic division $$ ( 3x^{3}+5x^{2}+12x+3 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}3&3&5&12&3\\& & 9& 42& \color{black}{162} \\ \hline &\color{blue}{3}&\color{blue}{14}&\color{blue}{54}&\color{orangered}{165} \end{array} $$The solution is:
$$ \frac{ 3x^{3}+5x^{2}+12x+3 }{ x-3 } = \color{blue}{3x^{2}+14x+54} ~+~ \frac{ \color{red}{ 165 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{3}&3&5&12&3\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}3&\color{orangered}{ 3 }&5&12&3\\& & & & \\ \hline &\color{orangered}{3}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&3&5&12&3\\& & \color{blue}{9} & & \\ \hline &\color{blue}{3}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ 9 } = \color{orangered}{ 14 } $
$$ \begin{array}{c|rrrr}3&3&\color{orangered}{ 5 }&12&3\\& & \color{orangered}{9} & & \\ \hline &3&\color{orangered}{14}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 14 } = \color{blue}{ 42 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&3&5&12&3\\& & 9& \color{blue}{42} & \\ \hline &3&\color{blue}{14}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ 42 } = \color{orangered}{ 54 } $
$$ \begin{array}{c|rrrr}3&3&5&\color{orangered}{ 12 }&3\\& & 9& \color{orangered}{42} & \\ \hline &3&14&\color{orangered}{54}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 54 } = \color{blue}{ 162 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&3&5&12&3\\& & 9& 42& \color{blue}{162} \\ \hline &3&14&\color{blue}{54}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 162 } = \color{orangered}{ 165 } $
$$ \begin{array}{c|rrrr}3&3&5&12&\color{orangered}{ 3 }\\& & 9& 42& \color{orangered}{162} \\ \hline &\color{blue}{3}&\color{blue}{14}&\color{blue}{54}&\color{orangered}{165} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{2}+14x+54 } $ with a remainder of $ \color{red}{ 165 } $.