Find remainder when $ 4x^{5}+1 $ is divided by $ x+3 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}-3&4&0&0&0&0&1\\& & -12& 36& -108& 324& \color{black}{-972} \\ \hline &\color{blue}{4}&\color{blue}{-12}&\color{blue}{36}&\color{blue}{-108}&\color{blue}{324}&\color{orangered}{-971} \end{array} $$The remainder when $ 4x^{5}+1 $ is divided by $ x+3 $ is $ \, \color{red}{ -971 } $.
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.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&4&0&0&0&0&1\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}-3&\color{orangered}{ 4 }&0&0&0&0&1\\& & & & & & \\ \hline &\color{orangered}{4}&&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 4 } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&4&0&0&0&0&1\\& & \color{blue}{-12} & & & & \\ \hline &\color{blue}{4}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ -12 } $
$$ \begin{array}{c|rrrrrr}-3&4&\color{orangered}{ 0 }&0&0&0&1\\& & \color{orangered}{-12} & & & & \\ \hline &4&\color{orangered}{-12}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -12 \right) } = \color{blue}{ 36 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&4&0&0&0&0&1\\& & -12& \color{blue}{36} & & & \\ \hline &4&\color{blue}{-12}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 36 } = \color{orangered}{ 36 } $
$$ \begin{array}{c|rrrrrr}-3&4&0&\color{orangered}{ 0 }&0&0&1\\& & -12& \color{orangered}{36} & & & \\ \hline &4&-12&\color{orangered}{36}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 36 } = \color{blue}{ -108 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&4&0&0&0&0&1\\& & -12& 36& \color{blue}{-108} & & \\ \hline &4&-12&\color{blue}{36}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -108 \right) } = \color{orangered}{ -108 } $
$$ \begin{array}{c|rrrrrr}-3&4&0&0&\color{orangered}{ 0 }&0&1\\& & -12& 36& \color{orangered}{-108} & & \\ \hline &4&-12&36&\color{orangered}{-108}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -108 \right) } = \color{blue}{ 324 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&4&0&0&0&0&1\\& & -12& 36& -108& \color{blue}{324} & \\ \hline &4&-12&36&\color{blue}{-108}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 324 } = \color{orangered}{ 324 } $
$$ \begin{array}{c|rrrrrr}-3&4&0&0&0&\color{orangered}{ 0 }&1\\& & -12& 36& -108& \color{orangered}{324} & \\ \hline &4&-12&36&-108&\color{orangered}{324}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 324 } = \color{blue}{ -972 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-3}&4&0&0&0&0&1\\& & -12& 36& -108& 324& \color{blue}{-972} \\ \hline &4&-12&36&-108&\color{blue}{324}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ \left( -972 \right) } = \color{orangered}{ -971 } $
$$ \begin{array}{c|rrrrrr}-3&4&0&0&0&0&\color{orangered}{ 1 }\\& & -12& 36& -108& 324& \color{orangered}{-972} \\ \hline &\color{blue}{4}&\color{blue}{-12}&\color{blue}{36}&\color{blue}{-108}&\color{blue}{324}&\color{orangered}{-971} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -971 }\right) $.