Find remainder when $ 4x^{4}+4x^{3}+25x-29 $ is divided by $ x+3 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-3&4&4&0&25&-29\\& & -12& 24& -72& \color{black}{141} \\ \hline &\color{blue}{4}&\color{blue}{-8}&\color{blue}{24}&\color{blue}{-47}&\color{orangered}{112} \end{array} $$The remainder when $ 4x^{4}+4x^{3}+25x-29 $ is divided by $ x+3 $ is $ \, \color{red}{ 112 } $.
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.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&4&4&0&25&-29\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-3&\color{orangered}{ 4 }&4&0&25&-29\\& & & & & \\ \hline &\color{orangered}{4}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&4&4&0&25&-29\\& & \color{blue}{-12} & & & \\ \hline &\color{blue}{4}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrrr}-3&4&\color{orangered}{ 4 }&0&25&-29\\& & \color{orangered}{-12} & & & \\ \hline &4&\color{orangered}{-8}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -8 \right) } = \color{blue}{ 24 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&4&4&0&25&-29\\& & -12& \color{blue}{24} & & \\ \hline &4&\color{blue}{-8}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 24 } = \color{orangered}{ 24 } $
$$ \begin{array}{c|rrrrr}-3&4&4&\color{orangered}{ 0 }&25&-29\\& & -12& \color{orangered}{24} & & \\ \hline &4&-8&\color{orangered}{24}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 24 } = \color{blue}{ -72 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&4&4&0&25&-29\\& & -12& 24& \color{blue}{-72} & \\ \hline &4&-8&\color{blue}{24}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 25 } + \color{orangered}{ \left( -72 \right) } = \color{orangered}{ -47 } $
$$ \begin{array}{c|rrrrr}-3&4&4&0&\color{orangered}{ 25 }&-29\\& & -12& 24& \color{orangered}{-72} & \\ \hline &4&-8&24&\color{orangered}{-47}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -47 \right) } = \color{blue}{ 141 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&4&4&0&25&-29\\& & -12& 24& -72& \color{blue}{141} \\ \hline &4&-8&24&\color{blue}{-47}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -29 } + \color{orangered}{ 141 } = \color{orangered}{ 112 } $
$$ \begin{array}{c|rrrrr}-3&4&4&0&25&\color{orangered}{ -29 }\\& & -12& 24& -72& \color{orangered}{141} \\ \hline &\color{blue}{4}&\color{blue}{-8}&\color{blue}{24}&\color{blue}{-47}&\color{orangered}{112} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 112 }\right) $.