Find remainder when $ x^{4}+x^{3}-7x^{2}+9x+4 $ is divided by $ x+4 $.
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-4&1&1&-7&9&4\\& & -4& 12& -20& \color{black}{44} \\ \hline &\color{blue}{1}&\color{blue}{-3}&\color{blue}{5}&\color{blue}{-11}&\color{orangered}{48} \end{array} $$The remainder when $ x^{4}+x^{3}-7x^{2}+9x+4 $ is divided by $ x+4 $ is $ \, \color{red}{ 48 } $.
We can find remainder using synthetic division method.
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.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&1&1&-7&9&4\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-4&\color{orangered}{ 1 }&1&-7&9&4\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 1 } = \color{blue}{ -4 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&1&1&-7&9&4\\& & \color{blue}{-4} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrrr}-4&1&\color{orangered}{ 1 }&-7&9&4\\& & \color{orangered}{-4} & & & \\ \hline &1&\color{orangered}{-3}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&1&1&-7&9&4\\& & -4& \color{blue}{12} & & \\ \hline &1&\color{blue}{-3}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 12 } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrrr}-4&1&1&\color{orangered}{ -7 }&9&4\\& & -4& \color{orangered}{12} & & \\ \hline &1&-3&\color{orangered}{5}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 5 } = \color{blue}{ -20 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&1&1&-7&9&4\\& & -4& 12& \color{blue}{-20} & \\ \hline &1&-3&\color{blue}{5}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 9 } + \color{orangered}{ \left( -20 \right) } = \color{orangered}{ -11 } $
$$ \begin{array}{c|rrrrr}-4&1&1&-7&\color{orangered}{ 9 }&4\\& & -4& 12& \color{orangered}{-20} & \\ \hline &1&-3&5&\color{orangered}{-11}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ \left( -11 \right) } = \color{blue}{ 44 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&1&1&-7&9&4\\& & -4& 12& -20& \color{blue}{44} \\ \hline &1&-3&5&\color{blue}{-11}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ 44 } = \color{orangered}{ 48 } $
$$ \begin{array}{c|rrrrr}-4&1&1&-7&9&\color{orangered}{ 4 }\\& & -4& 12& -20& \color{orangered}{44} \\ \hline &\color{blue}{1}&\color{blue}{-3}&\color{blue}{5}&\color{blue}{-11}&\color{orangered}{48} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 48 }\right) $.