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