Find remainder when $ x^{5}+9x^{4}+12x^{3}-26x^{2}-21x+16 $ is divided by $ x+3 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}-3&1&9&12&-26&-21&16\\& & -3& -18& 18& 24& \color{black}{-9} \\ \hline &\color{blue}{1}&\color{blue}{6}&\color{blue}{-6}&\color{blue}{-8}&\color{blue}{3}&\color{orangered}{7} \end{array} $$The remainder when $ x^{5}+9x^{4}+12x^{3}-26x^{2}-21x+16 $ is divided by $ x+3 $ is $ \, \color{red}{ 7 } $.
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 + 3 = 0 $ ( $ x = \color{blue}{ -3 } $ ) at the left.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&1&9&12&-26&-21&16\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}-3&\color{orangered}{ 1 }&9&12&-26&-21&16\\& & & & & & \\ \hline &\color{orangered}{1}&&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 1 } = \color{blue}{ -3 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&1&9&12&-26&-21&16\\& & \color{blue}{-3} & & & & \\ \hline &\color{blue}{1}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 9 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrrrr}-3&1&\color{orangered}{ 9 }&12&-26&-21&16\\& & \color{orangered}{-3} & & & & \\ \hline &1&\color{orangered}{6}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 6 } = \color{blue}{ -18 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&1&9&12&-26&-21&16\\& & -3& \color{blue}{-18} & & & \\ \hline &1&\color{blue}{6}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ \left( -18 \right) } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrrrr}-3&1&9&\color{orangered}{ 12 }&-26&-21&16\\& & -3& \color{orangered}{-18} & & & \\ \hline &1&6&\color{orangered}{-6}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -6 \right) } = \color{blue}{ 18 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&1&9&12&-26&-21&16\\& & -3& -18& \color{blue}{18} & & \\ \hline &1&6&\color{blue}{-6}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -26 } + \color{orangered}{ 18 } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrrrr}-3&1&9&12&\color{orangered}{ -26 }&-21&16\\& & -3& -18& \color{orangered}{18} & & \\ \hline &1&6&-6&\color{orangered}{-8}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -8 \right) } = \color{blue}{ 24 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&1&9&12&-26&-21&16\\& & -3& -18& 18& \color{blue}{24} & \\ \hline &1&6&-6&\color{blue}{-8}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -21 } + \color{orangered}{ 24 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrrrr}-3&1&9&12&-26&\color{orangered}{ -21 }&16\\& & -3& -18& 18& \color{orangered}{24} & \\ \hline &1&6&-6&-8&\color{orangered}{3}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 3 } = \color{blue}{ -9 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&1&9&12&-26&-21&16\\& & -3& -18& 18& 24& \color{blue}{-9} \\ \hline &1&6&-6&-8&\color{blue}{3}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ 16 } + \color{orangered}{ \left( -9 \right) } = \color{orangered}{ 7 } $
$$ \begin{array}{c|rrrrrr}-3&1&9&12&-26&-21&\color{orangered}{ 16 }\\& & -3& -18& 18& 24& \color{orangered}{-9} \\ \hline &\color{blue}{1}&\color{blue}{6}&\color{blue}{-6}&\color{blue}{-8}&\color{blue}{3}&\color{orangered}{7} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 7 }\right) $.