Check if $ x-3 $ is a factor of $ x^{2}-4x+3 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrr}3&1&-4&3\\& & 3& \color{black}{-3} \\ \hline &\color{blue}{1}&\color{blue}{-1}&\color{orangered}{0} \end{array} $$Because the remainder equals zero, we conclude that the $ x-3 $ is a factor of the $ x^{2}-4x+3 $.
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 -3 = 0 $ ( $ x = \color{blue}{ 3 } $ ) at the left.
$$ \begin{array}{c|rrr}\color{blue}{3}&1&-4&3\\& & & \\ \hline &&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrr}3&\color{orangered}{ 1 }&-4&3\\& & & \\ \hline &\color{orangered}{1}&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 1 } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrr}\color{blue}{3}&1&-4&3\\& & \color{blue}{3} & \\ \hline &\color{blue}{1}&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ 3 } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrr}3&1&\color{orangered}{ -4 }&3\\& & \color{orangered}{3} & \\ \hline &1&\color{orangered}{-1}& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ -3 } $.
$$ \begin{array}{c|rrr}\color{blue}{3}&1&-4&3\\& & 3& \color{blue}{-3} \\ \hline &1&\color{blue}{-1}& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrr}3&1&-4&\color{orangered}{ 3 }\\& & 3& \color{orangered}{-3} \\ \hline &\color{blue}{1}&\color{blue}{-1}&\color{orangered}{0} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 0 }\right)$.