Divide using synthetic division $$ ( 16x^{4}+21x^{3}+6x^{2}+6x+4 ) \div ( 4x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-2&16&21&6&6&4\\& & -32& 22& -56& \color{black}{100} \\ \hline &\color{blue}{16}&\color{blue}{-11}&\color{blue}{28}&\color{blue}{-50}&\color{orangered}{104} \end{array} $$The solution is:
$$ \frac{ 16x^{4}+21x^{3}+6x^{2}+6x+4 }{ x+2 } = \color{blue}{16x^{3}-11x^{2}+28x-50} ~+~ \frac{ \color{red}{ 104 } }{ x+2 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 2 = 0 $ ( $ x = \color{blue}{ -2 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&16&21&6&6&4\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-2&\color{orangered}{ 16 }&21&6&6&4\\& & & & & \\ \hline &\color{orangered}{16}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 16 } = \color{blue}{ -32 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&16&21&6&6&4\\& & \color{blue}{-32} & & & \\ \hline &\color{blue}{16}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 21 } + \color{orangered}{ \left( -32 \right) } = \color{orangered}{ -11 } $
$$ \begin{array}{c|rrrrr}-2&16&\color{orangered}{ 21 }&6&6&4\\& & \color{orangered}{-32} & & & \\ \hline &16&\color{orangered}{-11}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -11 \right) } = \color{blue}{ 22 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&16&21&6&6&4\\& & -32& \color{blue}{22} & & \\ \hline &16&\color{blue}{-11}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ 22 } = \color{orangered}{ 28 } $
$$ \begin{array}{c|rrrrr}-2&16&21&\color{orangered}{ 6 }&6&4\\& & -32& \color{orangered}{22} & & \\ \hline &16&-11&\color{orangered}{28}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 28 } = \color{blue}{ -56 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&16&21&6&6&4\\& & -32& 22& \color{blue}{-56} & \\ \hline &16&-11&\color{blue}{28}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ \left( -56 \right) } = \color{orangered}{ -50 } $
$$ \begin{array}{c|rrrrr}-2&16&21&6&\color{orangered}{ 6 }&4\\& & -32& 22& \color{orangered}{-56} & \\ \hline &16&-11&28&\color{orangered}{-50}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -50 \right) } = \color{blue}{ 100 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&16&21&6&6&4\\& & -32& 22& -56& \color{blue}{100} \\ \hline &16&-11&28&\color{blue}{-50}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ 100 } = \color{orangered}{ 104 } $
$$ \begin{array}{c|rrrrr}-2&16&21&6&6&\color{orangered}{ 4 }\\& & -32& 22& -56& \color{orangered}{100} \\ \hline &\color{blue}{16}&\color{blue}{-11}&\color{blue}{28}&\color{blue}{-50}&\color{orangered}{104} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 16x^{3}-11x^{2}+28x-50 } $ with a remainder of $ \color{red}{ 104 } $.