Divide using synthetic division $$ ( 5x^{4}-3x+1 ) \div ( -x+4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-4&5&0&0&-3&1\\& & -20& 80& -320& \color{black}{1292} \\ \hline &\color{blue}{5}&\color{blue}{-20}&\color{blue}{80}&\color{blue}{-323}&\color{orangered}{1293} \end{array} $$The solution is:
$$ \frac{ 5x^{4}-3x+1 }{ x+4 } = \color{blue}{5x^{3}-20x^{2}+80x-323} ~+~ \frac{ \color{red}{ 1293 } }{ x+4 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&5&0&0&-3&1\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-4&\color{orangered}{ 5 }&0&0&-3&1\\& & & & & \\ \hline &\color{orangered}{5}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 5 } = \color{blue}{ -20 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&5&0&0&-3&1\\& & \color{blue}{-20} & & & \\ \hline &\color{blue}{5}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -20 \right) } = \color{orangered}{ -20 } $
$$ \begin{array}{c|rrrrr}-4&5&\color{orangered}{ 0 }&0&-3&1\\& & \color{orangered}{-20} & & & \\ \hline &5&\color{orangered}{-20}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ \left( -20 \right) } = \color{blue}{ 80 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&5&0&0&-3&1\\& & -20& \color{blue}{80} & & \\ \hline &5&\color{blue}{-20}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 80 } = \color{orangered}{ 80 } $
$$ \begin{array}{c|rrrrr}-4&5&0&\color{orangered}{ 0 }&-3&1\\& & -20& \color{orangered}{80} & & \\ \hline &5&-20&\color{orangered}{80}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 80 } = \color{blue}{ -320 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&5&0&0&-3&1\\& & -20& 80& \color{blue}{-320} & \\ \hline &5&-20&\color{blue}{80}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ \left( -320 \right) } = \color{orangered}{ -323 } $
$$ \begin{array}{c|rrrrr}-4&5&0&0&\color{orangered}{ -3 }&1\\& & -20& 80& \color{orangered}{-320} & \\ \hline &5&-20&80&\color{orangered}{-323}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ \left( -323 \right) } = \color{blue}{ 1292 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&5&0&0&-3&1\\& & -20& 80& -320& \color{blue}{1292} \\ \hline &5&-20&80&\color{blue}{-323}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 1292 } = \color{orangered}{ 1293 } $
$$ \begin{array}{c|rrrrr}-4&5&0&0&-3&\color{orangered}{ 1 }\\& & -20& 80& -320& \color{orangered}{1292} \\ \hline &\color{blue}{5}&\color{blue}{-20}&\color{blue}{80}&\color{blue}{-323}&\color{orangered}{1293} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 5x^{3}-20x^{2}+80x-323 } $ with a remainder of $ \color{red}{ 1293 } $.