Find remainder when $ 6x^{6}+12x^{5}-2x^{3}-4x^{2}+5x+10 $ is divided by $ x+2 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrrr}-2&6&12&0&-2&-4&5&10\\& & -12& 0& 0& 4& 0& \color{black}{-10} \\ \hline &\color{blue}{6}&\color{blue}{0}&\color{blue}{0}&\color{blue}{-2}&\color{blue}{0}&\color{blue}{5}&\color{orangered}{0} \end{array} $$The remainder when $ 6x^{6}+12x^{5}-2x^{3}-4x^{2}+5x+10 $ is divided by $ x+2 $ is $ \, \color{red}{ 0 } $.
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.
$$ \begin{array}{c|rrrrrrr}\color{blue}{-2}&6&12&0&-2&-4&5&10\\& & & & & & & \\ \hline &&&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrrr}-2&\color{orangered}{ 6 }&12&0&-2&-4&5&10\\& & & & & & & \\ \hline &\color{orangered}{6}&&&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 6 } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrrrrr}\color{blue}{-2}&6&12&0&-2&-4&5&10\\& & \color{blue}{-12} & & & & & \\ \hline &\color{blue}{6}&&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrrrr}-2&6&\color{orangered}{ 12 }&0&-2&-4&5&10\\& & \color{orangered}{-12} & & & & & \\ \hline &6&\color{orangered}{0}&&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrrrr}\color{blue}{-2}&6&12&0&-2&-4&5&10\\& & -12& \color{blue}{0} & & & & \\ \hline &6&\color{blue}{0}&&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 0 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrrrr}-2&6&12&\color{orangered}{ 0 }&-2&-4&5&10\\& & -12& \color{orangered}{0} & & & & \\ \hline &6&0&\color{orangered}{0}&&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrrrr}\color{blue}{-2}&6&12&0&-2&-4&5&10\\& & -12& 0& \color{blue}{0} & & & \\ \hline &6&0&\color{blue}{0}&&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ 0 } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrrrrr}-2&6&12&0&\color{orangered}{ -2 }&-4&5&10\\& & -12& 0& \color{orangered}{0} & & & \\ \hline &6&0&0&\color{orangered}{-2}&&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrrrrr}\color{blue}{-2}&6&12&0&-2&-4&5&10\\& & -12& 0& 0& \color{blue}{4} & & \\ \hline &6&0&0&\color{blue}{-2}&&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ 4 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrrrr}-2&6&12&0&-2&\color{orangered}{ -4 }&5&10\\& & -12& 0& 0& \color{orangered}{4} & & \\ \hline &6&0&0&-2&\color{orangered}{0}&& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrrrr}\color{blue}{-2}&6&12&0&-2&-4&5&10\\& & -12& 0& 0& 4& \color{blue}{0} & \\ \hline &6&0&0&-2&\color{blue}{0}&& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ 0 } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrrrrr}-2&6&12&0&-2&-4&\color{orangered}{ 5 }&10\\& & -12& 0& 0& 4& \color{orangered}{0} & \\ \hline &6&0&0&-2&0&\color{orangered}{5}& \end{array} $$Step 12 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 5 } = \color{blue}{ -10 } $.
$$ \begin{array}{c|rrrrrrr}\color{blue}{-2}&6&12&0&-2&-4&5&10\\& & -12& 0& 0& 4& 0& \color{blue}{-10} \\ \hline &6&0&0&-2&0&\color{blue}{5}& \end{array} $$Step 13 : Add down last column: $ \color{orangered}{ 10 } + \color{orangered}{ \left( -10 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrrrr}-2&6&12&0&-2&-4&5&\color{orangered}{ 10 }\\& & -12& 0& 0& 4& 0& \color{orangered}{-10} \\ \hline &\color{blue}{6}&\color{blue}{0}&\color{blue}{0}&\color{blue}{-2}&\color{blue}{0}&\color{blue}{5}&\color{orangered}{0} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 0 }\right) $.