Check if $ x-1 $ is a factor of $ x^{3}+4x^{2}-19x+14 $.

The synthetic division table is:

$$ \begin{array}{c|rrrr}1&1&4&-19&14\\& & 1& 5& \color{black}{-14} \\ \hline &\color{blue}{1}&\color{blue}{5}&\color{blue}{-14}&\color{orangered}{0} \end{array} $$

Because the remainder equals zero, we conclude that the $ x-1 $ is a factor of the $ x^{3}+4x^{2}-19x+14 $.

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.

$$ \begin{array}{c|rrrr}\color{blue}{1}&1&4&-19&14\\& & & & \\ \hline &&&& \end{array} $$

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

$$ \begin{array}{c|rrrr}1&\color{orangered}{ 1 }&4&-19&14\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$

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

$$ \begin{array}{c|rrrr}\color{blue}{1}&1&4&-19&14\\& & \color{blue}{1} & & \\ \hline &\color{blue}{1}&&& \end{array} $$

Step 3 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ 1 } = \color{orangered}{ 5 } $

$$ \begin{array}{c|rrrr}1&1&\color{orangered}{ 4 }&-19&14\\& & \color{orangered}{1} & & \\ \hline &1&\color{orangered}{5}&& \end{array} $$

Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 5 } = \color{blue}{ 5 } $.

$$ \begin{array}{c|rrrr}\color{blue}{1}&1&4&-19&14\\& & 1& \color{blue}{5} & \\ \hline &1&\color{blue}{5}&& \end{array} $$

Step 5 : Add down last column: $ \color{orangered}{ -19 } + \color{orangered}{ 5 } = \color{orangered}{ -14 } $

$$ \begin{array}{c|rrrr}1&1&4&\color{orangered}{ -19 }&14\\& & 1& \color{orangered}{5} & \\ \hline &1&5&\color{orangered}{-14}& \end{array} $$

Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -14 \right) } = \color{blue}{ -14 } $.

$$ \begin{array}{c|rrrr}\color{blue}{1}&1&4&-19&14\\& & 1& 5& \color{blue}{-14} \\ \hline &1&5&\color{blue}{-14}& \end{array} $$

Step 7 : Add down last column: $ \color{orangered}{ 14 } + \color{orangered}{ \left( -14 \right) } = \color{orangered}{ 0 } $

$$ \begin{array}{c|rrrr}1&1&4&-19&\color{orangered}{ 14 }\\& & 1& 5& \color{orangered}{-14} \\ \hline &\color{blue}{1}&\color{blue}{5}&\color{blue}{-14}&\color{orangered}{0} \end{array} $$

Remainder is the last entry in the bottom row $ \left(\color{red}{ 0 }\right)$.

Synthetic Division Calculator