Divide using synthetic division $$ ( 7x^{4}+3x^{3}+2x+1 ) \div ( x+4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-4&7&3&0&2&1\\& & -28& 100& -400& \color{black}{1592} \\ \hline &\color{blue}{7}&\color{blue}{-25}&\color{blue}{100}&\color{blue}{-398}&\color{orangered}{1593} \end{array} $$The solution is:
$$ \frac{ 7x^{4}+3x^{3}+2x+1 }{ x+4 } = \color{blue}{7x^{3}-25x^{2}+100x-398} ~+~ \frac{ \color{red}{ 1593 } }{ x+4 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&7&3&0&2&1\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-4&\color{orangered}{ 7 }&3&0&2&1\\& & & & & \\ \hline &\color{orangered}{7}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 7 } = \color{blue}{ -28 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&7&3&0&2&1\\& & \color{blue}{-28} & & & \\ \hline &\color{blue}{7}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ \left( -28 \right) } = \color{orangered}{ -25 } $
$$ \begin{array}{c|rrrrr}-4&7&\color{orangered}{ 3 }&0&2&1\\& & \color{orangered}{-28} & & & \\ \hline &7&\color{orangered}{-25}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ \left( -25 \right) } = \color{blue}{ 100 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&7&3&0&2&1\\& & -28& \color{blue}{100} & & \\ \hline &7&\color{blue}{-25}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 100 } = \color{orangered}{ 100 } $
$$ \begin{array}{c|rrrrr}-4&7&3&\color{orangered}{ 0 }&2&1\\& & -28& \color{orangered}{100} & & \\ \hline &7&-25&\color{orangered}{100}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 100 } = \color{blue}{ -400 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&7&3&0&2&1\\& & -28& 100& \color{blue}{-400} & \\ \hline &7&-25&\color{blue}{100}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ \left( -400 \right) } = \color{orangered}{ -398 } $
$$ \begin{array}{c|rrrrr}-4&7&3&0&\color{orangered}{ 2 }&1\\& & -28& 100& \color{orangered}{-400} & \\ \hline &7&-25&100&\color{orangered}{-398}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ \left( -398 \right) } = \color{blue}{ 1592 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&7&3&0&2&1\\& & -28& 100& -400& \color{blue}{1592} \\ \hline &7&-25&100&\color{blue}{-398}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 1592 } = \color{orangered}{ 1593 } $
$$ \begin{array}{c|rrrrr}-4&7&3&0&2&\color{orangered}{ 1 }\\& & -28& 100& -400& \color{orangered}{1592} \\ \hline &\color{blue}{7}&\color{blue}{-25}&\color{blue}{100}&\color{blue}{-398}&\color{orangered}{1593} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 7x^{3}-25x^{2}+100x-398 } $ with a remainder of $ \color{red}{ 1593 } $.