Divide using synthetic division $$ ( 5x^{3}+20x^{2}-2x+4 ) \div ( x+4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-4&5&20&-2&4\\& & -20& 0& \color{black}{8} \\ \hline &\color{blue}{5}&\color{blue}{0}&\color{blue}{-2}&\color{orangered}{12} \end{array} $$The solution is:
$$ \frac{ 5x^{3}+20x^{2}-2x+4 }{ x+4 } = \color{blue}{5x^{2}-2} ~+~ \frac{ \color{red}{ 12 } }{ x+4 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&5&20&-2&4\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-4&\color{orangered}{ 5 }&20&-2&4\\& & & & \\ \hline &\color{orangered}{5}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 5 } = \color{blue}{ -20 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&5&20&-2&4\\& & \color{blue}{-20} & & \\ \hline &\color{blue}{5}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 20 } + \color{orangered}{ \left( -20 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}-4&5&\color{orangered}{ 20 }&-2&4\\& & \color{orangered}{-20} & & \\ \hline &5&\color{orangered}{0}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&5&20&-2&4\\& & -20& \color{blue}{0} & \\ \hline &5&\color{blue}{0}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ 0 } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrr}-4&5&20&\color{orangered}{ -2 }&4\\& & -20& \color{orangered}{0} & \\ \hline &5&0&\color{orangered}{-2}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ 8 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&5&20&-2&4\\& & -20& 0& \color{blue}{8} \\ \hline &5&0&\color{blue}{-2}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ 8 } = \color{orangered}{ 12 } $
$$ \begin{array}{c|rrrr}-4&5&20&-2&\color{orangered}{ 4 }\\& & -20& 0& \color{orangered}{8} \\ \hline &\color{blue}{5}&\color{blue}{0}&\color{blue}{-2}&\color{orangered}{12} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 5x^{2}-2 } $ with a remainder of $ \color{red}{ 12 } $.