Divide using synthetic division $$ ( 5x^{5}+19x^{4}-2x^{3}+7x^{2}-11x-28 ) \div ( x+4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}-4&5&19&-2&7&-11&-28\\& & -20& 4& -8& 4& \color{black}{28} \\ \hline &\color{blue}{5}&\color{blue}{-1}&\color{blue}{2}&\color{blue}{-1}&\color{blue}{-7}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 5x^{5}+19x^{4}-2x^{3}+7x^{2}-11x-28 }{ x+4 } = \color{blue}{5x^{4}-x^{3}+2x^{2}-x-7} $$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.
$$ \begin{array}{c|rrrrrr}\color{blue}{-4}&5&19&-2&7&-11&-28\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}-4&\color{orangered}{ 5 }&19&-2&7&-11&-28\\& & & & & & \\ \hline &\color{orangered}{5}&&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-4}&5&19&-2&7&-11&-28\\& & \color{blue}{-20} & & & & \\ \hline &\color{blue}{5}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 19 } + \color{orangered}{ \left( -20 \right) } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrrrr}-4&5&\color{orangered}{ 19 }&-2&7&-11&-28\\& & \color{orangered}{-20} & & & & \\ \hline &5&\color{orangered}{-1}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-4}&5&19&-2&7&-11&-28\\& & -20& \color{blue}{4} & & & \\ \hline &5&\color{blue}{-1}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ 4 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrrr}-4&5&19&\color{orangered}{ -2 }&7&-11&-28\\& & -20& \color{orangered}{4} & & & \\ \hline &5&-1&\color{orangered}{2}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 2 } = \color{blue}{ -8 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-4}&5&19&-2&7&-11&-28\\& & -20& 4& \color{blue}{-8} & & \\ \hline &5&-1&\color{blue}{2}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ \left( -8 \right) } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrrrr}-4&5&19&-2&\color{orangered}{ 7 }&-11&-28\\& & -20& 4& \color{orangered}{-8} & & \\ \hline &5&-1&2&\color{orangered}{-1}&& \end{array} $$Step 8 : 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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-4}&5&19&-2&7&-11&-28\\& & -20& 4& -8& \color{blue}{4} & \\ \hline &5&-1&2&\color{blue}{-1}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -11 } + \color{orangered}{ 4 } = \color{orangered}{ -7 } $
$$ \begin{array}{c|rrrrrr}-4&5&19&-2&7&\color{orangered}{ -11 }&-28\\& & -20& 4& -8& \color{orangered}{4} & \\ \hline &5&-1&2&-1&\color{orangered}{-7}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ \left( -7 \right) } = \color{blue}{ 28 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-4}&5&19&-2&7&-11&-28\\& & -20& 4& -8& 4& \color{blue}{28} \\ \hline &5&-1&2&-1&\color{blue}{-7}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ -28 } + \color{orangered}{ 28 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrrr}-4&5&19&-2&7&-11&\color{orangered}{ -28 }\\& & -20& 4& -8& 4& \color{orangered}{28} \\ \hline &\color{blue}{5}&\color{blue}{-1}&\color{blue}{2}&\color{blue}{-1}&\color{blue}{-7}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 5x^{4}-x^{3}+2x^{2}-x-7 } $ with a remainder of $ \color{red}{ 0 } $.