Find remainder when $ x^{4}-12x^{2}-67 $ is divided by $ x+4 $.
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-4&1&0&-12&0&-67\\& & -4& 16& -16& \color{black}{64} \\ \hline &\color{blue}{1}&\color{blue}{-4}&\color{blue}{4}&\color{blue}{-16}&\color{orangered}{-3} \end{array} $$The remainder when $ x^{4}-12x^{2}-67 $ is divided by $ x+4 $ is $ \, \color{red}{ -3 } $.
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.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&1&0&-12&0&-67\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-4&\color{orangered}{ 1 }&0&-12&0&-67\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&1&0&-12&0&-67\\& & \color{blue}{-4} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrrr}-4&1&\color{orangered}{ 0 }&-12&0&-67\\& & \color{orangered}{-4} & & & \\ \hline &1&\color{orangered}{-4}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ 16 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&1&0&-12&0&-67\\& & -4& \color{blue}{16} & & \\ \hline &1&\color{blue}{-4}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -12 } + \color{orangered}{ 16 } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrrr}-4&1&0&\color{orangered}{ -12 }&0&-67\\& & -4& \color{orangered}{16} & & \\ \hline &1&-4&\color{orangered}{4}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 4 } = \color{blue}{ -16 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&1&0&-12&0&-67\\& & -4& 16& \color{blue}{-16} & \\ \hline &1&-4&\color{blue}{4}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -16 \right) } = \color{orangered}{ -16 } $
$$ \begin{array}{c|rrrrr}-4&1&0&-12&\color{orangered}{ 0 }&-67\\& & -4& 16& \color{orangered}{-16} & \\ \hline &1&-4&4&\color{orangered}{-16}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ \left( -16 \right) } = \color{blue}{ 64 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&1&0&-12&0&-67\\& & -4& 16& -16& \color{blue}{64} \\ \hline &1&-4&4&\color{blue}{-16}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -67 } + \color{orangered}{ 64 } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrrr}-4&1&0&-12&0&\color{orangered}{ -67 }\\& & -4& 16& -16& \color{orangered}{64} \\ \hline &\color{blue}{1}&\color{blue}{-4}&\color{blue}{4}&\color{blue}{-16}&\color{orangered}{-3} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -3 }\right) $.