Find remainder when $ x^{5}+x^{2}-10x+113 $ is divided by $ x+2 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}-2&1&0&0&1&-10&113\\& & -2& 4& -8& 14& \color{black}{-8} \\ \hline &\color{blue}{1}&\color{blue}{-2}&\color{blue}{4}&\color{blue}{-7}&\color{blue}{4}&\color{orangered}{105} \end{array} $$The remainder when $ x^{5}+x^{2}-10x+113 $ is divided by $ x+2 $ is $ \, \color{red}{ 105 } $.
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.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&1&0&0&1&-10&113\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}-2&\color{orangered}{ 1 }&0&0&1&-10&113\\& & & & & & \\ \hline &\color{orangered}{1}&&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&1&0&0&1&-10&113\\& & \color{blue}{-2} & & & & \\ \hline &\color{blue}{1}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -2 \right) } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrrrr}-2&1&\color{orangered}{ 0 }&0&1&-10&113\\& & \color{orangered}{-2} & & & & \\ \hline &1&\color{orangered}{-2}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&1&0&0&1&-10&113\\& & -2& \color{blue}{4} & & & \\ \hline &1&\color{blue}{-2}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 4 } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrrrr}-2&1&0&\color{orangered}{ 0 }&1&-10&113\\& & -2& \color{orangered}{4} & & & \\ \hline &1&-2&\color{orangered}{4}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 4 } = \color{blue}{ -8 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&1&0&0&1&-10&113\\& & -2& 4& \color{blue}{-8} & & \\ \hline &1&-2&\color{blue}{4}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ \left( -8 \right) } = \color{orangered}{ -7 } $
$$ \begin{array}{c|rrrrrr}-2&1&0&0&\color{orangered}{ 1 }&-10&113\\& & -2& 4& \color{orangered}{-8} & & \\ \hline &1&-2&4&\color{orangered}{-7}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -7 \right) } = \color{blue}{ 14 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&1&0&0&1&-10&113\\& & -2& 4& -8& \color{blue}{14} & \\ \hline &1&-2&4&\color{blue}{-7}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -10 } + \color{orangered}{ 14 } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrrrr}-2&1&0&0&1&\color{orangered}{ -10 }&113\\& & -2& 4& -8& \color{orangered}{14} & \\ \hline &1&-2&4&-7&\color{orangered}{4}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 4 } = \color{blue}{ -8 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&1&0&0&1&-10&113\\& & -2& 4& -8& 14& \color{blue}{-8} \\ \hline &1&-2&4&-7&\color{blue}{4}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ 113 } + \color{orangered}{ \left( -8 \right) } = \color{orangered}{ 105 } $
$$ \begin{array}{c|rrrrrr}-2&1&0&0&1&-10&\color{orangered}{ 113 }\\& & -2& 4& -8& 14& \color{orangered}{-8} \\ \hline &\color{blue}{1}&\color{blue}{-2}&\color{blue}{4}&\color{blue}{-7}&\color{blue}{4}&\color{orangered}{105} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 105 }\right) $.