Divide using synthetic division $$ ( x^{3}+8x^{2}+19x+12 ) \div ( x-4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}4&1&8&19&12\\& & 4& 48& \color{black}{268} \\ \hline &\color{blue}{1}&\color{blue}{12}&\color{blue}{67}&\color{orangered}{280} \end{array} $$The solution is:
$$ \frac{ x^{3}+8x^{2}+19x+12 }{ x-4 } = \color{blue}{x^{2}+12x+67} ~+~ \frac{ \color{red}{ 280 } }{ x-4 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{4}&1&8&19&12\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}4&\color{orangered}{ 1 }&8&19&12\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&1&8&19&12\\& & \color{blue}{4} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 8 } + \color{orangered}{ 4 } = \color{orangered}{ 12 } $
$$ \begin{array}{c|rrrr}4&1&\color{orangered}{ 8 }&19&12\\& & \color{orangered}{4} & & \\ \hline &1&\color{orangered}{12}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 12 } = \color{blue}{ 48 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&1&8&19&12\\& & 4& \color{blue}{48} & \\ \hline &1&\color{blue}{12}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 19 } + \color{orangered}{ 48 } = \color{orangered}{ 67 } $
$$ \begin{array}{c|rrrr}4&1&8&\color{orangered}{ 19 }&12\\& & 4& \color{orangered}{48} & \\ \hline &1&12&\color{orangered}{67}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 67 } = \color{blue}{ 268 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&1&8&19&12\\& & 4& 48& \color{blue}{268} \\ \hline &1&12&\color{blue}{67}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ 268 } = \color{orangered}{ 280 } $
$$ \begin{array}{c|rrrr}4&1&8&19&\color{orangered}{ 12 }\\& & 4& 48& \color{orangered}{268} \\ \hline &\color{blue}{1}&\color{blue}{12}&\color{blue}{67}&\color{orangered}{280} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}+12x+67 } $ with a remainder of $ \color{red}{ 280 } $.