Divide using synthetic division $$ ( 5x^{3}+21x^{2}-16 ) \div ( x+4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-4&5&21&0&-16\\& & -20& -4& \color{black}{16} \\ \hline &\color{blue}{5}&\color{blue}{1}&\color{blue}{-4}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 5x^{3}+21x^{2}-16 }{ x+4 } = \color{blue}{5x^{2}+x-4} $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&5&21&0&-16\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-4&\color{orangered}{ 5 }&21&0&-16\\& & & & \\ \hline &\color{orangered}{5}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&5&21&0&-16\\& & \color{blue}{-20} & & \\ \hline &\color{blue}{5}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 21 } + \color{orangered}{ \left( -20 \right) } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrr}-4&5&\color{orangered}{ 21 }&0&-16\\& & \color{orangered}{-20} & & \\ \hline &5&\color{orangered}{1}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 1 } = \color{blue}{ -4 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&5&21&0&-16\\& & -20& \color{blue}{-4} & \\ \hline &5&\color{blue}{1}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrr}-4&5&21&\color{orangered}{ 0 }&-16\\& & -20& \color{orangered}{-4} & \\ \hline &5&1&\color{orangered}{-4}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ 16 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-4}&5&21&0&-16\\& & -20& -4& \color{blue}{16} \\ \hline &5&1&\color{blue}{-4}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -16 } + \color{orangered}{ 16 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}-4&5&21&0&\color{orangered}{ -16 }\\& & -20& -4& \color{orangered}{16} \\ \hline &\color{blue}{5}&\color{blue}{1}&\color{blue}{-4}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 5x^{2}+x-4 } $ with a remainder of $ \color{red}{ 0 } $.