Find remainder when $ -4x^{4}+x^{3}-4x^{2}+1 $ is divided by $ x+5 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-5&-4&1&-4&0&1\\& & 20& -105& 545& \color{black}{-2725} \\ \hline &\color{blue}{-4}&\color{blue}{21}&\color{blue}{-109}&\color{blue}{545}&\color{orangered}{-2724} \end{array} $$The remainder when $ -4x^{4}+x^{3}-4x^{2}+1 $ is divided by $ x+5 $ is $ \, \color{red}{ -2724 } $.
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 + 5 = 0 $ ( $ x = \color{blue}{ -5 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{-5}&-4&1&-4&0&1\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-5&\color{orangered}{ -4 }&1&-4&0&1\\& & & & & \\ \hline &\color{orangered}{-4}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ 20 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-5}&-4&1&-4&0&1\\& & \color{blue}{20} & & & \\ \hline &\color{blue}{-4}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 20 } = \color{orangered}{ 21 } $
$$ \begin{array}{c|rrrrr}-5&-4&\color{orangered}{ 1 }&-4&0&1\\& & \color{orangered}{20} & & & \\ \hline &-4&\color{orangered}{21}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 21 } = \color{blue}{ -105 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-5}&-4&1&-4&0&1\\& & 20& \color{blue}{-105} & & \\ \hline &-4&\color{blue}{21}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ \left( -105 \right) } = \color{orangered}{ -109 } $
$$ \begin{array}{c|rrrrr}-5&-4&1&\color{orangered}{ -4 }&0&1\\& & 20& \color{orangered}{-105} & & \\ \hline &-4&21&\color{orangered}{-109}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ \left( -109 \right) } = \color{blue}{ 545 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-5}&-4&1&-4&0&1\\& & 20& -105& \color{blue}{545} & \\ \hline &-4&21&\color{blue}{-109}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 545 } = \color{orangered}{ 545 } $
$$ \begin{array}{c|rrrrr}-5&-4&1&-4&\color{orangered}{ 0 }&1\\& & 20& -105& \color{orangered}{545} & \\ \hline &-4&21&-109&\color{orangered}{545}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 545 } = \color{blue}{ -2725 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-5}&-4&1&-4&0&1\\& & 20& -105& 545& \color{blue}{-2725} \\ \hline &-4&21&-109&\color{blue}{545}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ \left( -2725 \right) } = \color{orangered}{ -2724 } $
$$ \begin{array}{c|rrrrr}-5&-4&1&-4&0&\color{orangered}{ 1 }\\& & 20& -105& 545& \color{orangered}{-2725} \\ \hline &\color{blue}{-4}&\color{blue}{21}&\color{blue}{-109}&\color{blue}{545}&\color{orangered}{-2724} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -2724 }\right) $.