Check if $ x-3 $ is a factor of $ 4x^{4}-2x^{3}+9x-6 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}3&4&-2&0&9&-6\\& & 12& 30& 90& \color{black}{297} \\ \hline &\color{blue}{4}&\color{blue}{10}&\color{blue}{30}&\color{blue}{99}&\color{orangered}{291} \end{array} $$Because the remainder $ \left( \color{red}{ 291 } \right) $ is not zero, we conclude that the $ x-3 $ is not a factor of $ 4x^{4}-2x^{3}+9x-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.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&4&-2&0&9&-6\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}3&\color{orangered}{ 4 }&-2&0&9&-6\\& & & & & \\ \hline &\color{orangered}{4}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 4 } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&4&-2&0&9&-6\\& & \color{blue}{12} & & & \\ \hline &\color{blue}{4}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ 12 } = \color{orangered}{ 10 } $
$$ \begin{array}{c|rrrrr}3&4&\color{orangered}{ -2 }&0&9&-6\\& & \color{orangered}{12} & & & \\ \hline &4&\color{orangered}{10}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 10 } = \color{blue}{ 30 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&4&-2&0&9&-6\\& & 12& \color{blue}{30} & & \\ \hline &4&\color{blue}{10}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 30 } = \color{orangered}{ 30 } $
$$ \begin{array}{c|rrrrr}3&4&-2&\color{orangered}{ 0 }&9&-6\\& & 12& \color{orangered}{30} & & \\ \hline &4&10&\color{orangered}{30}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 30 } = \color{blue}{ 90 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&4&-2&0&9&-6\\& & 12& 30& \color{blue}{90} & \\ \hline &4&10&\color{blue}{30}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 9 } + \color{orangered}{ 90 } = \color{orangered}{ 99 } $
$$ \begin{array}{c|rrrrr}3&4&-2&0&\color{orangered}{ 9 }&-6\\& & 12& 30& \color{orangered}{90} & \\ \hline &4&10&30&\color{orangered}{99}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 99 } = \color{blue}{ 297 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&4&-2&0&9&-6\\& & 12& 30& 90& \color{blue}{297} \\ \hline &4&10&30&\color{blue}{99}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ 297 } = \color{orangered}{ 291 } $
$$ \begin{array}{c|rrrrr}3&4&-2&0&9&\color{orangered}{ -6 }\\& & 12& 30& 90& \color{orangered}{297} \\ \hline &\color{blue}{4}&\color{blue}{10}&\color{blue}{30}&\color{blue}{99}&\color{orangered}{291} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 291 }\right)$.