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