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