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