Check if $ x-3 $ is a factor of $ x^{4}+18x^{2}+81 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}3&1&0&18&0&81\\& & 3& 9& 81& \color{black}{243} \\ \hline &\color{blue}{1}&\color{blue}{3}&\color{blue}{27}&\color{blue}{81}&\color{orangered}{324} \end{array} $$Because the remainder $ \left( \color{red}{ 324 } \right) $ is not zero, we conclude that the $ x-3 $ is not a factor of $ x^{4}+18x^{2}+81$.
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.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&0&18&0&81\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}3&\color{orangered}{ 1 }&0&18&0&81\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&0&18&0&81\\& & \color{blue}{3} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 3 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrrr}3&1&\color{orangered}{ 0 }&18&0&81\\& & \color{orangered}{3} & & & \\ \hline &1&\color{orangered}{3}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 3 } = \color{blue}{ 9 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&0&18&0&81\\& & 3& \color{blue}{9} & & \\ \hline &1&\color{blue}{3}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 18 } + \color{orangered}{ 9 } = \color{orangered}{ 27 } $
$$ \begin{array}{c|rrrrr}3&1&0&\color{orangered}{ 18 }&0&81\\& & 3& \color{orangered}{9} & & \\ \hline &1&3&\color{orangered}{27}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 27 } = \color{blue}{ 81 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&0&18&0&81\\& & 3& 9& \color{blue}{81} & \\ \hline &1&3&\color{blue}{27}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 81 } = \color{orangered}{ 81 } $
$$ \begin{array}{c|rrrrr}3&1&0&18&\color{orangered}{ 0 }&81\\& & 3& 9& \color{orangered}{81} & \\ \hline &1&3&27&\color{orangered}{81}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 81 } = \color{blue}{ 243 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&0&18&0&81\\& & 3& 9& 81& \color{blue}{243} \\ \hline &1&3&27&\color{blue}{81}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 81 } + \color{orangered}{ 243 } = \color{orangered}{ 324 } $
$$ \begin{array}{c|rrrrr}3&1&0&18&0&\color{orangered}{ 81 }\\& & 3& 9& 81& \color{orangered}{243} \\ \hline &\color{blue}{1}&\color{blue}{3}&\color{blue}{27}&\color{blue}{81}&\color{orangered}{324} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 324 }\right)$.