Check if $ x-3 $ is a factor of $ 3x^{2}-7x-6 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrr}3&3&-7&-6\\& & 9& \color{black}{6} \\ \hline &\color{blue}{3}&\color{blue}{2}&\color{orangered}{0} \end{array} $$Because the remainder equals zero, we conclude that the $ x-3 $ is a factor of the $ 3x^{2}-7x-6 $.
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.
$$ \begin{array}{c|rrr}\color{blue}{3}&3&-7&-6\\& & & \\ \hline &&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrr}3&\color{orangered}{ 3 }&-7&-6\\& & & \\ \hline &\color{orangered}{3}&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 3 } = \color{blue}{ 9 } $.
$$ \begin{array}{c|rrr}\color{blue}{3}&3&-7&-6\\& & \color{blue}{9} & \\ \hline &\color{blue}{3}&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 9 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrr}3&3&\color{orangered}{ -7 }&-6\\& & \color{orangered}{9} & \\ \hline &3&\color{orangered}{2}& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 2 } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrr}\color{blue}{3}&3&-7&-6\\& & 9& \color{blue}{6} \\ \hline &3&\color{blue}{2}& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ 6 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrr}3&3&-7&\color{orangered}{ -6 }\\& & 9& \color{orangered}{6} \\ \hline &\color{blue}{3}&\color{blue}{2}&\color{orangered}{0} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 0 }\right)$.