Find remainder when $ x^{5}-5x^{3}+2x $ is divided by $ x+3 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}-3&1&0&-5&0&2&0\\& & -3& 9& -12& 36& \color{black}{-114} \\ \hline &\color{blue}{1}&\color{blue}{-3}&\color{blue}{4}&\color{blue}{-12}&\color{blue}{38}&\color{orangered}{-114} \end{array} $$The remainder when $ x^{5}-5x^{3}+2x $ is divided by $ x+3 $ is $ \, \color{red}{ -114 } $.
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.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&1&0&-5&0&2&0\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}-3&\color{orangered}{ 1 }&0&-5&0&2&0\\& & & & & & \\ \hline &\color{orangered}{1}&&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&1&0&-5&0&2&0\\& & \color{blue}{-3} & & & & \\ \hline &\color{blue}{1}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrrrr}-3&1&\color{orangered}{ 0 }&-5&0&2&0\\& & \color{orangered}{-3} & & & & \\ \hline &1&\color{orangered}{-3}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ 9 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&1&0&-5&0&2&0\\& & -3& \color{blue}{9} & & & \\ \hline &1&\color{blue}{-3}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ 9 } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrrrr}-3&1&0&\color{orangered}{ -5 }&0&2&0\\& & -3& \color{orangered}{9} & & & \\ \hline &1&-3&\color{orangered}{4}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 4 } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&1&0&-5&0&2&0\\& & -3& 9& \color{blue}{-12} & & \\ \hline &1&-3&\color{blue}{4}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ -12 } $
$$ \begin{array}{c|rrrrrr}-3&1&0&-5&\color{orangered}{ 0 }&2&0\\& & -3& 9& \color{orangered}{-12} & & \\ \hline &1&-3&4&\color{orangered}{-12}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -12 \right) } = \color{blue}{ 36 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&1&0&-5&0&2&0\\& & -3& 9& -12& \color{blue}{36} & \\ \hline &1&-3&4&\color{blue}{-12}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 36 } = \color{orangered}{ 38 } $
$$ \begin{array}{c|rrrrrr}-3&1&0&-5&0&\color{orangered}{ 2 }&0\\& & -3& 9& -12& \color{orangered}{36} & \\ \hline &1&-3&4&-12&\color{orangered}{38}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 38 } = \color{blue}{ -114 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&1&0&-5&0&2&0\\& & -3& 9& -12& 36& \color{blue}{-114} \\ \hline &1&-3&4&-12&\color{blue}{38}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -114 \right) } = \color{orangered}{ -114 } $
$$ \begin{array}{c|rrrrrr}-3&1&0&-5&0&2&\color{orangered}{ 0 }\\& & -3& 9& -12& 36& \color{orangered}{-114} \\ \hline &\color{blue}{1}&\color{blue}{-3}&\color{blue}{4}&\color{blue}{-12}&\color{blue}{38}&\color{orangered}{-114} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -114 }\right) $.