Check if $ x+4 $ is a factor of $ x^{4}-4x^{3}+3x^{2}-3x-57 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-4&1&-4&3&-3&-57\\& & -4& 32& -140& \color{black}{572} \\ \hline &\color{blue}{1}&\color{blue}{-8}&\color{blue}{35}&\color{blue}{-143}&\color{orangered}{515} \end{array} $$Because the remainder $ \left( \color{red}{ 515 } \right) $ is not zero, we conclude that the $ x+4 $ is not a factor of $ x^{4}-4x^{3}+3x^{2}-3x-57$.
Explanation
First we need to create a synthetic division table.
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 4 = 0 $ ( $ x = \color{blue}{ -4 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&1&-4&3&-3&-57\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-4&\color{orangered}{ 1 }&-4&3&-3&-57\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 1 } = \color{blue}{ -4 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&1&-4&3&-3&-57\\& & \color{blue}{-4} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrrr}-4&1&\color{orangered}{ -4 }&3&-3&-57\\& & \color{orangered}{-4} & & & \\ \hline &1&\color{orangered}{-8}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ \left( -8 \right) } = \color{blue}{ 32 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&1&-4&3&-3&-57\\& & -4& \color{blue}{32} & & \\ \hline &1&\color{blue}{-8}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 32 } = \color{orangered}{ 35 } $
$$ \begin{array}{c|rrrrr}-4&1&-4&\color{orangered}{ 3 }&-3&-57\\& & -4& \color{orangered}{32} & & \\ \hline &1&-8&\color{orangered}{35}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 35 } = \color{blue}{ -140 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&1&-4&3&-3&-57\\& & -4& 32& \color{blue}{-140} & \\ \hline &1&-8&\color{blue}{35}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ \left( -140 \right) } = \color{orangered}{ -143 } $
$$ \begin{array}{c|rrrrr}-4&1&-4&3&\color{orangered}{ -3 }&-57\\& & -4& 32& \color{orangered}{-140} & \\ \hline &1&-8&35&\color{orangered}{-143}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ \left( -143 \right) } = \color{blue}{ 572 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&1&-4&3&-3&-57\\& & -4& 32& -140& \color{blue}{572} \\ \hline &1&-8&35&\color{blue}{-143}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -57 } + \color{orangered}{ 572 } = \color{orangered}{ 515 } $
$$ \begin{array}{c|rrrrr}-4&1&-4&3&-3&\color{orangered}{ -57 }\\& & -4& 32& -140& \color{orangered}{572} \\ \hline &\color{blue}{1}&\color{blue}{-8}&\color{blue}{35}&\color{blue}{-143}&\color{orangered}{515} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 515 }\right)$.