Check if $ x+5 $ is a factor of $ x^{3}-4x^{2}+3x+7 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-5&1&-4&3&7\\& & -5& 45& \color{black}{-240} \\ \hline &\color{blue}{1}&\color{blue}{-9}&\color{blue}{48}&\color{orangered}{-233} \end{array} $$Because the remainder $ \left( \color{red}{ -233 } \right) $ is not zero, we conclude that the $ x+5 $ is not a factor of $ x^{3}-4x^{2}+3x+7$.
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 + 5 = 0 $ ( $ x = \color{blue}{ -5 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&1&-4&3&7\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-5&\color{orangered}{ 1 }&-4&3&7\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 1 } = \color{blue}{ -5 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&1&-4&3&7\\& & \color{blue}{-5} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ \left( -5 \right) } = \color{orangered}{ -9 } $
$$ \begin{array}{c|rrrr}-5&1&\color{orangered}{ -4 }&3&7\\& & \color{orangered}{-5} & & \\ \hline &1&\color{orangered}{-9}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ \left( -9 \right) } = \color{blue}{ 45 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&1&-4&3&7\\& & -5& \color{blue}{45} & \\ \hline &1&\color{blue}{-9}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 45 } = \color{orangered}{ 48 } $
$$ \begin{array}{c|rrrr}-5&1&-4&\color{orangered}{ 3 }&7\\& & -5& \color{orangered}{45} & \\ \hline &1&-9&\color{orangered}{48}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 48 } = \color{blue}{ -240 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&1&-4&3&7\\& & -5& 45& \color{blue}{-240} \\ \hline &1&-9&\color{blue}{48}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ \left( -240 \right) } = \color{orangered}{ -233 } $
$$ \begin{array}{c|rrrr}-5&1&-4&3&\color{orangered}{ 7 }\\& & -5& 45& \color{orangered}{-240} \\ \hline &\color{blue}{1}&\color{blue}{-9}&\color{blue}{48}&\color{orangered}{-233} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -233 }\right)$.