Check if $ x-5 $ is a factor of $ x^{3}-4x^{2}+3x+7 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}5&1&-4&3&7\\& & 5& 5& \color{black}{40} \\ \hline &\color{blue}{1}&\color{blue}{1}&\color{blue}{8}&\color{orangered}{47} \end{array} $$Because the remainder $ \left( \color{red}{ 47 } \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}{ 5 } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrr}5&1&\color{orangered}{ -4 }&3&7\\& & \color{orangered}{5} & & \\ \hline &1&\color{orangered}{1}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 1 } = \color{blue}{ 5 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&1&-4&3&7\\& & 5& \color{blue}{5} & \\ \hline &1&\color{blue}{1}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 5 } = \color{orangered}{ 8 } $
$$ \begin{array}{c|rrrr}5&1&-4&\color{orangered}{ 3 }&7\\& & 5& \color{orangered}{5} & \\ \hline &1&1&\color{orangered}{8}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 8 } = \color{blue}{ 40 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&1&-4&3&7\\& & 5& 5& \color{blue}{40} \\ \hline &1&1&\color{blue}{8}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ 40 } = \color{orangered}{ 47 } $
$$ \begin{array}{c|rrrr}5&1&-4&3&\color{orangered}{ 7 }\\& & 5& 5& \color{orangered}{40} \\ \hline &\color{blue}{1}&\color{blue}{1}&\color{blue}{8}&\color{orangered}{47} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 47 }\right)$.