Divide using synthetic division $$ ( x^{3}+3x^{2}+12 ) \div ( x+4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-4&1&3&0&12\\& & -4& 4& \color{black}{-16} \\ \hline &\color{blue}{1}&\color{blue}{-1}&\color{blue}{4}&\color{orangered}{-4} \end{array} $$The solution is:
$$ \frac{ x^{3}+3x^{2}+12 }{ x+4 } = \color{blue}{x^{2}-x+4} \color{red}{~-~} \frac{ \color{red}{ 4 } }{ x+4 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 4 = 0 $ ( $ x = \color{blue}{ -4 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&1&3&0&12\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-4&\color{orangered}{ 1 }&3&0&12\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 1 } = \color{blue}{ -4 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&1&3&0&12\\& & \color{blue}{-4} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrr}-4&1&\color{orangered}{ 3 }&0&12\\& & \color{orangered}{-4} & & \\ \hline &1&\color{orangered}{-1}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&1&3&0&12\\& & -4& \color{blue}{4} & \\ \hline &1&\color{blue}{-1}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 4 } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrr}-4&1&3&\color{orangered}{ 0 }&12\\& & -4& \color{orangered}{4} & \\ \hline &1&-1&\color{orangered}{4}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 4 } = \color{blue}{ -16 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&1&3&0&12\\& & -4& 4& \color{blue}{-16} \\ \hline &1&-1&\color{blue}{4}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ \left( -16 \right) } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrr}-4&1&3&0&\color{orangered}{ 12 }\\& & -4& 4& \color{orangered}{-16} \\ \hline &\color{blue}{1}&\color{blue}{-1}&\color{blue}{4}&\color{orangered}{-4} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}-x+4 } $ with a remainder of $ \color{red}{ -4 } $.