Find remainder when $ 5x^{5}+24x^{4}+23x^{3}+37x+20 $ is divided by $ x+3 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}-3&5&24&23&0&37&20\\& & -15& -27& 12& -36& \color{black}{-3} \\ \hline &\color{blue}{5}&\color{blue}{9}&\color{blue}{-4}&\color{blue}{12}&\color{blue}{1}&\color{orangered}{17} \end{array} $$The remainder when $ 5x^{5}+24x^{4}+23x^{3}+37x+20 $ is divided by $ x+3 $ is $ \, \color{red}{ 17 } $.
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.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&5&24&23&0&37&20\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}-3&\color{orangered}{ 5 }&24&23&0&37&20\\& & & & & & \\ \hline &\color{orangered}{5}&&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 5 } = \color{blue}{ -15 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&5&24&23&0&37&20\\& & \color{blue}{-15} & & & & \\ \hline &\color{blue}{5}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 24 } + \color{orangered}{ \left( -15 \right) } = \color{orangered}{ 9 } $
$$ \begin{array}{c|rrrrrr}-3&5&\color{orangered}{ 24 }&23&0&37&20\\& & \color{orangered}{-15} & & & & \\ \hline &5&\color{orangered}{9}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 9 } = \color{blue}{ -27 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&5&24&23&0&37&20\\& & -15& \color{blue}{-27} & & & \\ \hline &5&\color{blue}{9}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 23 } + \color{orangered}{ \left( -27 \right) } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrrrr}-3&5&24&\color{orangered}{ 23 }&0&37&20\\& & -15& \color{orangered}{-27} & & & \\ \hline &5&9&\color{orangered}{-4}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&5&24&23&0&37&20\\& & -15& -27& \color{blue}{12} & & \\ \hline &5&9&\color{blue}{-4}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 12 } = \color{orangered}{ 12 } $
$$ \begin{array}{c|rrrrrr}-3&5&24&23&\color{orangered}{ 0 }&37&20\\& & -15& -27& \color{orangered}{12} & & \\ \hline &5&9&-4&\color{orangered}{12}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 12 } = \color{blue}{ -36 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&5&24&23&0&37&20\\& & -15& -27& 12& \color{blue}{-36} & \\ \hline &5&9&-4&\color{blue}{12}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 37 } + \color{orangered}{ \left( -36 \right) } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrrrr}-3&5&24&23&0&\color{orangered}{ 37 }&20\\& & -15& -27& 12& \color{orangered}{-36} & \\ \hline &5&9&-4&12&\color{orangered}{1}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 1 } = \color{blue}{ -3 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&5&24&23&0&37&20\\& & -15& -27& 12& -36& \color{blue}{-3} \\ \hline &5&9&-4&12&\color{blue}{1}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ 20 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ 17 } $
$$ \begin{array}{c|rrrrrr}-3&5&24&23&0&37&\color{orangered}{ 20 }\\& & -15& -27& 12& -36& \color{orangered}{-3} \\ \hline &\color{blue}{5}&\color{blue}{9}&\color{blue}{-4}&\color{blue}{12}&\color{blue}{1}&\color{orangered}{17} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 17 }\right) $.