Find remainder when $ 7x^{5}-6x^{4}-3x+9 $ is divided by $ x+2 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}-2&7&-6&0&0&-3&9\\& & -14& 40& -80& 160& \color{black}{-314} \\ \hline &\color{blue}{7}&\color{blue}{-20}&\color{blue}{40}&\color{blue}{-80}&\color{blue}{157}&\color{orangered}{-305} \end{array} $$The remainder when $ 7x^{5}-6x^{4}-3x+9 $ is divided by $ x+2 $ is $ \, \color{red}{ -305 } $.
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.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&7&-6&0&0&-3&9\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}-2&\color{orangered}{ 7 }&-6&0&0&-3&9\\& & & & & & \\ \hline &\color{orangered}{7}&&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 7 } = \color{blue}{ -14 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&7&-6&0&0&-3&9\\& & \color{blue}{-14} & & & & \\ \hline &\color{blue}{7}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ \left( -14 \right) } = \color{orangered}{ -20 } $
$$ \begin{array}{c|rrrrrr}-2&7&\color{orangered}{ -6 }&0&0&-3&9\\& & \color{orangered}{-14} & & & & \\ \hline &7&\color{orangered}{-20}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -20 \right) } = \color{blue}{ 40 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&7&-6&0&0&-3&9\\& & -14& \color{blue}{40} & & & \\ \hline &7&\color{blue}{-20}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 40 } = \color{orangered}{ 40 } $
$$ \begin{array}{c|rrrrrr}-2&7&-6&\color{orangered}{ 0 }&0&-3&9\\& & -14& \color{orangered}{40} & & & \\ \hline &7&-20&\color{orangered}{40}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 40 } = \color{blue}{ -80 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&7&-6&0&0&-3&9\\& & -14& 40& \color{blue}{-80} & & \\ \hline &7&-20&\color{blue}{40}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -80 \right) } = \color{orangered}{ -80 } $
$$ \begin{array}{c|rrrrrr}-2&7&-6&0&\color{orangered}{ 0 }&-3&9\\& & -14& 40& \color{orangered}{-80} & & \\ \hline &7&-20&40&\color{orangered}{-80}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -80 \right) } = \color{blue}{ 160 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&7&-6&0&0&-3&9\\& & -14& 40& -80& \color{blue}{160} & \\ \hline &7&-20&40&\color{blue}{-80}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ 160 } = \color{orangered}{ 157 } $
$$ \begin{array}{c|rrrrrr}-2&7&-6&0&0&\color{orangered}{ -3 }&9\\& & -14& 40& -80& \color{orangered}{160} & \\ \hline &7&-20&40&-80&\color{orangered}{157}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 157 } = \color{blue}{ -314 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&7&-6&0&0&-3&9\\& & -14& 40& -80& 160& \color{blue}{-314} \\ \hline &7&-20&40&-80&\color{blue}{157}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ 9 } + \color{orangered}{ \left( -314 \right) } = \color{orangered}{ -305 } $
$$ \begin{array}{c|rrrrrr}-2&7&-6&0&0&-3&\color{orangered}{ 9 }\\& & -14& 40& -80& 160& \color{orangered}{-314} \\ \hline &\color{blue}{7}&\color{blue}{-20}&\color{blue}{40}&\color{blue}{-80}&\color{blue}{157}&\color{orangered}{-305} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -305 }\right) $.