Find remainder when $ 6x^{4}+25x^{3}+20x^{2}-9x-9 $ is divided by $ x+2 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-2&6&25&20&-9&-9\\& & -12& -26& 12& \color{black}{-6} \\ \hline &\color{blue}{6}&\color{blue}{13}&\color{blue}{-6}&\color{blue}{3}&\color{orangered}{-15} \end{array} $$The remainder when $ 6x^{4}+25x^{3}+20x^{2}-9x-9 $ is divided by $ x+2 $ is $ \, \color{red}{ -15 } $.
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.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&6&25&20&-9&-9\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-2&\color{orangered}{ 6 }&25&20&-9&-9\\& & & & & \\ \hline &\color{orangered}{6}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&6&25&20&-9&-9\\& & \color{blue}{-12} & & & \\ \hline &\color{blue}{6}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 25 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ 13 } $
$$ \begin{array}{c|rrrrr}-2&6&\color{orangered}{ 25 }&20&-9&-9\\& & \color{orangered}{-12} & & & \\ \hline &6&\color{orangered}{13}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 13 } = \color{blue}{ -26 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&6&25&20&-9&-9\\& & -12& \color{blue}{-26} & & \\ \hline &6&\color{blue}{13}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 20 } + \color{orangered}{ \left( -26 \right) } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrrr}-2&6&25&\color{orangered}{ 20 }&-9&-9\\& & -12& \color{orangered}{-26} & & \\ \hline &6&13&\color{orangered}{-6}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -6 \right) } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&6&25&20&-9&-9\\& & -12& -26& \color{blue}{12} & \\ \hline &6&13&\color{blue}{-6}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -9 } + \color{orangered}{ 12 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrrr}-2&6&25&20&\color{orangered}{ -9 }&-9\\& & -12& -26& \color{orangered}{12} & \\ \hline &6&13&-6&\color{orangered}{3}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 3 } = \color{blue}{ -6 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&6&25&20&-9&-9\\& & -12& -26& 12& \color{blue}{-6} \\ \hline &6&13&-6&\color{blue}{3}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -9 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ -15 } $
$$ \begin{array}{c|rrrrr}-2&6&25&20&-9&\color{orangered}{ -9 }\\& & -12& -26& 12& \color{orangered}{-6} \\ \hline &\color{blue}{6}&\color{blue}{13}&\color{blue}{-6}&\color{blue}{3}&\color{orangered}{-15} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -15 }\right) $.