Find remainder when $ x^{5}+7x^{4}-35x^{2}+2x+2 $ is divided by $ x+3 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}-3&1&7&0&-35&2&2\\& & -3& -12& 36& -3& \color{black}{3} \\ \hline &\color{blue}{1}&\color{blue}{4}&\color{blue}{-12}&\color{blue}{1}&\color{blue}{-1}&\color{orangered}{5} \end{array} $$The remainder when $ x^{5}+7x^{4}-35x^{2}+2x+2 $ is divided by $ x+3 $ is $ \, \color{red}{ 5 } $.
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.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&1&7&0&-35&2&2\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}-3&\color{orangered}{ 1 }&7&0&-35&2&2\\& & & & & & \\ \hline &\color{orangered}{1}&&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&1&7&0&-35&2&2\\& & \color{blue}{-3} & & & & \\ \hline &\color{blue}{1}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrrrr}-3&1&\color{orangered}{ 7 }&0&-35&2&2\\& & \color{orangered}{-3} & & & & \\ \hline &1&\color{orangered}{4}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 4 } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&1&7&0&-35&2&2\\& & -3& \color{blue}{-12} & & & \\ \hline &1&\color{blue}{4}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ -12 } $
$$ \begin{array}{c|rrrrrr}-3&1&7&\color{orangered}{ 0 }&-35&2&2\\& & -3& \color{orangered}{-12} & & & \\ \hline &1&4&\color{orangered}{-12}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -12 \right) } = \color{blue}{ 36 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&1&7&0&-35&2&2\\& & -3& -12& \color{blue}{36} & & \\ \hline &1&4&\color{blue}{-12}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -35 } + \color{orangered}{ 36 } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrrrr}-3&1&7&0&\color{orangered}{ -35 }&2&2\\& & -3& -12& \color{orangered}{36} & & \\ \hline &1&4&-12&\color{orangered}{1}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 1 } = \color{blue}{ -3 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&1&7&0&-35&2&2\\& & -3& -12& 36& \color{blue}{-3} & \\ \hline &1&4&-12&\color{blue}{1}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrrrr}-3&1&7&0&-35&\color{orangered}{ 2 }&2\\& & -3& -12& 36& \color{orangered}{-3} & \\ \hline &1&4&-12&1&\color{orangered}{-1}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&1&7&0&-35&2&2\\& & -3& -12& 36& -3& \color{blue}{3} \\ \hline &1&4&-12&1&\color{blue}{-1}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 3 } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrrrr}-3&1&7&0&-35&2&\color{orangered}{ 2 }\\& & -3& -12& 36& -3& \color{orangered}{3} \\ \hline &\color{blue}{1}&\color{blue}{4}&\color{blue}{-12}&\color{blue}{1}&\color{blue}{-1}&\color{orangered}{5} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 5 }\right) $.