Find remainder when $ x^{5}+3x^{4}+2x^{3}+3x^{2}-2x+1 $ is divided by $ x+3 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}-3&1&3&2&3&-2&1\\& & -3& 0& -6& 9& \color{black}{-21} \\ \hline &\color{blue}{1}&\color{blue}{0}&\color{blue}{2}&\color{blue}{-3}&\color{blue}{7}&\color{orangered}{-20} \end{array} $$The remainder when $ x^{5}+3x^{4}+2x^{3}+3x^{2}-2x+1 $ is divided by $ x+3 $ is $ \, \color{red}{ -20 } $.
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.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&1&3&2&3&-2&1\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}-3&\color{orangered}{ 1 }&3&2&3&-2&1\\& & & & & & \\ \hline &\color{orangered}{1}&&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&1&3&2&3&-2&1\\& & \color{blue}{-3} & & & & \\ \hline &\color{blue}{1}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrrr}-3&1&\color{orangered}{ 3 }&2&3&-2&1\\& & \color{orangered}{-3} & & & & \\ \hline &1&\color{orangered}{0}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&1&3&2&3&-2&1\\& & -3& \color{blue}{0} & & & \\ \hline &1&\color{blue}{0}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 0 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrrr}-3&1&3&\color{orangered}{ 2 }&3&-2&1\\& & -3& \color{orangered}{0} & & & \\ \hline &1&0&\color{orangered}{2}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 2 } = \color{blue}{ -6 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&1&3&2&3&-2&1\\& & -3& 0& \color{blue}{-6} & & \\ \hline &1&0&\color{blue}{2}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrrrr}-3&1&3&2&\color{orangered}{ 3 }&-2&1\\& & -3& 0& \color{orangered}{-6} & & \\ \hline &1&0&2&\color{orangered}{-3}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ 9 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&1&3&2&3&-2&1\\& & -3& 0& -6& \color{blue}{9} & \\ \hline &1&0&2&\color{blue}{-3}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ 9 } = \color{orangered}{ 7 } $
$$ \begin{array}{c|rrrrrr}-3&1&3&2&3&\color{orangered}{ -2 }&1\\& & -3& 0& -6& \color{orangered}{9} & \\ \hline &1&0&2&-3&\color{orangered}{7}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 7 } = \color{blue}{ -21 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&1&3&2&3&-2&1\\& & -3& 0& -6& 9& \color{blue}{-21} \\ \hline &1&0&2&-3&\color{blue}{7}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ \left( -21 \right) } = \color{orangered}{ -20 } $
$$ \begin{array}{c|rrrrrr}-3&1&3&2&3&-2&\color{orangered}{ 1 }\\& & -3& 0& -6& 9& \color{orangered}{-21} \\ \hline &\color{blue}{1}&\color{blue}{0}&\color{blue}{2}&\color{blue}{-3}&\color{blue}{7}&\color{orangered}{-20} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -20 }\right) $.