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