Check if $ x-1 $ is a factor of $ 3x^{4}-14x^{2}-5 $.

The synthetic division table is:

$$ \begin{array}{c|rrrrr}1&3&0&-14&0&-5\\& & 3& 3& -11& \color{black}{-11} \\ \hline &\color{blue}{3}&\color{blue}{3}&\color{blue}{-11}&\color{blue}{-11}&\color{orangered}{-16} \end{array} $$

Because the remainder $ \left( \color{red}{ -16 } \right) $ is not zero, we conclude that the $ x-1 $ is not a factor of $ 3x^{4}-14x^{2}-5$.

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 -1 = 0 $ ( $ x = \color{blue}{ 1 } $ ) at the left.

$$ \begin{array}{c|rrrrr}\color{blue}{1}&3&0&-14&0&-5\\& & & & & \\ \hline &&&&& \end{array} $$

Step 1 : Bring down the leading coefficient to the bottom row.

$$ \begin{array}{c|rrrrr}1&\color{orangered}{ 3 }&0&-14&0&-5\\& & & & & \\ \hline &\color{orangered}{3}&&&& \end{array} $$

Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 3 } = \color{blue}{ 3 } $.

$$ \begin{array}{c|rrrrr}\color{blue}{1}&3&0&-14&0&-5\\& & \color{blue}{3} & & & \\ \hline &\color{blue}{3}&&&& \end{array} $$

Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 3 } = \color{orangered}{ 3 } $

$$ \begin{array}{c|rrrrr}1&3&\color{orangered}{ 0 }&-14&0&-5\\& & \color{orangered}{3} & & & \\ \hline &3&\color{orangered}{3}&&& \end{array} $$

Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 3 } = \color{blue}{ 3 } $.

$$ \begin{array}{c|rrrrr}\color{blue}{1}&3&0&-14&0&-5\\& & 3& \color{blue}{3} & & \\ \hline &3&\color{blue}{3}&&& \end{array} $$

Step 5 : Add down last column: $ \color{orangered}{ -14 } + \color{orangered}{ 3 } = \color{orangered}{ -11 } $

$$ \begin{array}{c|rrrrr}1&3&0&\color{orangered}{ -14 }&0&-5\\& & 3& \color{orangered}{3} & & \\ \hline &3&3&\color{orangered}{-11}&& \end{array} $$

Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -11 \right) } = \color{blue}{ -11 } $.

$$ \begin{array}{c|rrrrr}\color{blue}{1}&3&0&-14&0&-5\\& & 3& 3& \color{blue}{-11} & \\ \hline &3&3&\color{blue}{-11}&& \end{array} $$

Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -11 \right) } = \color{orangered}{ -11 } $

$$ \begin{array}{c|rrrrr}1&3&0&-14&\color{orangered}{ 0 }&-5\\& & 3& 3& \color{orangered}{-11} & \\ \hline &3&3&-11&\color{orangered}{-11}& \end{array} $$

Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -11 \right) } = \color{blue}{ -11 } $.

$$ \begin{array}{c|rrrrr}\color{blue}{1}&3&0&-14&0&-5\\& & 3& 3& -11& \color{blue}{-11} \\ \hline &3&3&-11&\color{blue}{-11}& \end{array} $$

Step 9 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ \left( -11 \right) } = \color{orangered}{ -16 } $

$$ \begin{array}{c|rrrrr}1&3&0&-14&0&\color{orangered}{ -5 }\\& & 3& 3& -11& \color{orangered}{-11} \\ \hline &\color{blue}{3}&\color{blue}{3}&\color{blue}{-11}&\color{blue}{-11}&\color{orangered}{-16} \end{array} $$

Remainder is the last entry in the bottom row $ \left(\color{red}{ -16 }\right)$.

Synthetic Division Calculator