Divide using synthetic division $$ ( 5x^{4}+16x^{3}-15x^{2}+8x+16 ) \div ( x+4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-4&5&16&-15&8&16\\& & -20& 16& -4& \color{black}{-16} \\ \hline &\color{blue}{5}&\color{blue}{-4}&\color{blue}{1}&\color{blue}{4}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 5x^{4}+16x^{3}-15x^{2}+8x+16 }{ x+4 } = \color{blue}{5x^{3}-4x^{2}+x+4} $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&5&16&-15&8&16\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-4&\color{orangered}{ 5 }&16&-15&8&16\\& & & & & \\ \hline &\color{orangered}{5}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&5&16&-15&8&16\\& & \color{blue}{-20} & & & \\ \hline &\color{blue}{5}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 16 } + \color{orangered}{ \left( -20 \right) } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrrr}-4&5&\color{orangered}{ 16 }&-15&8&16\\& & \color{orangered}{-20} & & & \\ \hline &5&\color{orangered}{-4}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ 16 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&5&16&-15&8&16\\& & -20& \color{blue}{16} & & \\ \hline &5&\color{blue}{-4}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -15 } + \color{orangered}{ 16 } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrrr}-4&5&16&\color{orangered}{ -15 }&8&16\\& & -20& \color{orangered}{16} & & \\ \hline &5&-4&\color{orangered}{1}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 1 } = \color{blue}{ -4 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&5&16&-15&8&16\\& & -20& 16& \color{blue}{-4} & \\ \hline &5&-4&\color{blue}{1}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 8 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrrr}-4&5&16&-15&\color{orangered}{ 8 }&16\\& & -20& 16& \color{orangered}{-4} & \\ \hline &5&-4&1&\color{orangered}{4}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 4 } = \color{blue}{ -16 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&5&16&-15&8&16\\& & -20& 16& -4& \color{blue}{-16} \\ \hline &5&-4&1&\color{blue}{4}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 16 } + \color{orangered}{ \left( -16 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}-4&5&16&-15&8&\color{orangered}{ 16 }\\& & -20& 16& -4& \color{orangered}{-16} \\ \hline &\color{blue}{5}&\color{blue}{-4}&\color{blue}{1}&\color{blue}{4}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 5x^{3}-4x^{2}+x+4 } $ with a remainder of $ \color{red}{ 0 } $.