Find remainder when $ x^{4}-x^{3}+2 $ is divided by $ x+2 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-2&1&-1&0&0&2\\& & -2& 6& -12& \color{black}{24} \\ \hline &\color{blue}{1}&\color{blue}{-3}&\color{blue}{6}&\color{blue}{-12}&\color{orangered}{26} \end{array} $$The remainder when $ x^{4}-x^{3}+2 $ is divided by $ x+2 $ is $ \, \color{red}{ 26 } $.
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 + 2 = 0 $ ( $ x = \color{blue}{ -2 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&1&-1&0&0&2\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-2&\color{orangered}{ 1 }&-1&0&0&2\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 1 } = \color{blue}{ -2 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&1&-1&0&0&2\\& & \color{blue}{-2} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ \left( -2 \right) } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrrr}-2&1&\color{orangered}{ -1 }&0&0&2\\& & \color{orangered}{-2} & & & \\ \hline &1&\color{orangered}{-3}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&1&-1&0&0&2\\& & -2& \color{blue}{6} & & \\ \hline &1&\color{blue}{-3}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 6 } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrrr}-2&1&-1&\color{orangered}{ 0 }&0&2\\& & -2& \color{orangered}{6} & & \\ \hline &1&-3&\color{orangered}{6}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 6 } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&1&-1&0&0&2\\& & -2& 6& \color{blue}{-12} & \\ \hline &1&-3&\color{blue}{6}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ -12 } $
$$ \begin{array}{c|rrrrr}-2&1&-1&0&\color{orangered}{ 0 }&2\\& & -2& 6& \color{orangered}{-12} & \\ \hline &1&-3&6&\color{orangered}{-12}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -12 \right) } = \color{blue}{ 24 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&1&-1&0&0&2\\& & -2& 6& -12& \color{blue}{24} \\ \hline &1&-3&6&\color{blue}{-12}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 24 } = \color{orangered}{ 26 } $
$$ \begin{array}{c|rrrrr}-2&1&-1&0&0&\color{orangered}{ 2 }\\& & -2& 6& -12& \color{orangered}{24} \\ \hline &\color{blue}{1}&\color{blue}{-3}&\color{blue}{6}&\color{blue}{-12}&\color{orangered}{26} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 26 }\right) $.