Divide using synthetic division $$ ( 5x^{5}+7x^{2}-11x-28 ) \div ( x+4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}-4&5&0&0&7&-11&-28\\& & -20& 80& -320& 1252& \color{black}{-4964} \\ \hline &\color{blue}{5}&\color{blue}{-20}&\color{blue}{80}&\color{blue}{-313}&\color{blue}{1241}&\color{orangered}{-4992} \end{array} $$The solution is:
$$ \frac{ 5x^{5}+7x^{2}-11x-28 }{ x+4 } = \color{blue}{5x^{4}-20x^{3}+80x^{2}-313x+1241} \color{red}{~-~} \frac{ \color{red}{ 4992 } }{ x+4 } $$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.
$$ \begin{array}{c|rrrrrr}\color{blue}{-4}&5&0&0&7&-11&-28\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}-4&\color{orangered}{ 5 }&0&0&7&-11&-28\\& & & & & & \\ \hline &\color{orangered}{5}&&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-4}&5&0&0&7&-11&-28\\& & \color{blue}{-20} & & & & \\ \hline &\color{blue}{5}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -20 \right) } = \color{orangered}{ -20 } $
$$ \begin{array}{c|rrrrrr}-4&5&\color{orangered}{ 0 }&0&7&-11&-28\\& & \color{orangered}{-20} & & & & \\ \hline &5&\color{orangered}{-20}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ \left( -20 \right) } = \color{blue}{ 80 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-4}&5&0&0&7&-11&-28\\& & -20& \color{blue}{80} & & & \\ \hline &5&\color{blue}{-20}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 80 } = \color{orangered}{ 80 } $
$$ \begin{array}{c|rrrrrr}-4&5&0&\color{orangered}{ 0 }&7&-11&-28\\& & -20& \color{orangered}{80} & & & \\ \hline &5&-20&\color{orangered}{80}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 80 } = \color{blue}{ -320 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-4}&5&0&0&7&-11&-28\\& & -20& 80& \color{blue}{-320} & & \\ \hline &5&-20&\color{blue}{80}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ \left( -320 \right) } = \color{orangered}{ -313 } $
$$ \begin{array}{c|rrrrrr}-4&5&0&0&\color{orangered}{ 7 }&-11&-28\\& & -20& 80& \color{orangered}{-320} & & \\ \hline &5&-20&80&\color{orangered}{-313}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ \left( -313 \right) } = \color{blue}{ 1252 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-4}&5&0&0&7&-11&-28\\& & -20& 80& -320& \color{blue}{1252} & \\ \hline &5&-20&80&\color{blue}{-313}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -11 } + \color{orangered}{ 1252 } = \color{orangered}{ 1241 } $
$$ \begin{array}{c|rrrrrr}-4&5&0&0&7&\color{orangered}{ -11 }&-28\\& & -20& 80& -320& \color{orangered}{1252} & \\ \hline &5&-20&80&-313&\color{orangered}{1241}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 1241 } = \color{blue}{ -4964 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-4}&5&0&0&7&-11&-28\\& & -20& 80& -320& 1252& \color{blue}{-4964} \\ \hline &5&-20&80&-313&\color{blue}{1241}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ -28 } + \color{orangered}{ \left( -4964 \right) } = \color{orangered}{ -4992 } $
$$ \begin{array}{c|rrrrrr}-4&5&0&0&7&-11&\color{orangered}{ -28 }\\& & -20& 80& -320& 1252& \color{orangered}{-4964} \\ \hline &\color{blue}{5}&\color{blue}{-20}&\color{blue}{80}&\color{blue}{-313}&\color{blue}{1241}&\color{orangered}{-4992} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 5x^{4}-20x^{3}+80x^{2}-313x+1241 } $ with a remainder of $ \color{red}{ -4992 } $.