Divide using synthetic division $$ ( 4x^{3}+20x^{2}+19x+18 ) \div ( x+4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-4&4&20&19&18\\& & -16& -16& \color{black}{-12} \\ \hline &\color{blue}{4}&\color{blue}{4}&\color{blue}{3}&\color{orangered}{6} \end{array} $$The solution is:
$$ \frac{ 4x^{3}+20x^{2}+19x+18 }{ x+4 } = \color{blue}{4x^{2}+4x+3} ~+~ \frac{ \color{red}{ 6 } }{ x+4 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&4&20&19&18\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-4&\color{orangered}{ 4 }&20&19&18\\& & & & \\ \hline &\color{orangered}{4}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 4 } = \color{blue}{ -16 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&4&20&19&18\\& & \color{blue}{-16} & & \\ \hline &\color{blue}{4}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 20 } + \color{orangered}{ \left( -16 \right) } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrr}-4&4&\color{orangered}{ 20 }&19&18\\& & \color{orangered}{-16} & & \\ \hline &4&\color{orangered}{4}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 4 } = \color{blue}{ -16 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&4&20&19&18\\& & -16& \color{blue}{-16} & \\ \hline &4&\color{blue}{4}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 19 } + \color{orangered}{ \left( -16 \right) } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrr}-4&4&20&\color{orangered}{ 19 }&18\\& & -16& \color{orangered}{-16} & \\ \hline &4&4&\color{orangered}{3}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 3 } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&4&20&19&18\\& & -16& -16& \color{blue}{-12} \\ \hline &4&4&\color{blue}{3}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 18 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrr}-4&4&20&19&\color{orangered}{ 18 }\\& & -16& -16& \color{orangered}{-12} \\ \hline &\color{blue}{4}&\color{blue}{4}&\color{blue}{3}&\color{orangered}{6} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{2}+4x+3 } $ with a remainder of $ \color{red}{ 6 } $.