Check if $ x-2 $ is a factor of $ x^{3}+3x^{2}-10x+24 $.

The synthetic division table is:

$$ \begin{array}{c|rrrr}2&1&3&-10&24\\& & 2& 10& \color{black}{0} \\ \hline &\color{blue}{1}&\color{blue}{5}&\color{blue}{0}&\color{orangered}{24} \end{array} $$

Because the remainder $ \left( \color{red}{ 24 } \right) $ is not zero, we conclude that the $ x-2 $ is not a factor of $ x^{3}+3x^{2}-10x+24$.

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 -2 = 0 $ ( $ x = \color{blue}{ 2 } $ ) at the left.

$$ \begin{array}{c|rrrr}\color{blue}{2}&1&3&-10&24\\& & & & \\ \hline &&&& \end{array} $$

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

$$ \begin{array}{c|rrrr}2&\color{orangered}{ 1 }&3&-10&24\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$

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

$$ \begin{array}{c|rrrr}\color{blue}{2}&1&3&-10&24\\& & \color{blue}{2} & & \\ \hline &\color{blue}{1}&&& \end{array} $$

Step 3 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 2 } = \color{orangered}{ 5 } $

$$ \begin{array}{c|rrrr}2&1&\color{orangered}{ 3 }&-10&24\\& & \color{orangered}{2} & & \\ \hline &1&\color{orangered}{5}&& \end{array} $$

Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 5 } = \color{blue}{ 10 } $.

$$ \begin{array}{c|rrrr}\color{blue}{2}&1&3&-10&24\\& & 2& \color{blue}{10} & \\ \hline &1&\color{blue}{5}&& \end{array} $$

Step 5 : Add down last column: $ \color{orangered}{ -10 } + \color{orangered}{ 10 } = \color{orangered}{ 0 } $

$$ \begin{array}{c|rrrr}2&1&3&\color{orangered}{ -10 }&24\\& & 2& \color{orangered}{10} & \\ \hline &1&5&\color{orangered}{0}& \end{array} $$

Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.

$$ \begin{array}{c|rrrr}\color{blue}{2}&1&3&-10&24\\& & 2& 10& \color{blue}{0} \\ \hline &1&5&\color{blue}{0}& \end{array} $$

Step 7 : Add down last column: $ \color{orangered}{ 24 } + \color{orangered}{ 0 } = \color{orangered}{ 24 } $

$$ \begin{array}{c|rrrr}2&1&3&-10&\color{orangered}{ 24 }\\& & 2& 10& \color{orangered}{0} \\ \hline &\color{blue}{1}&\color{blue}{5}&\color{blue}{0}&\color{orangered}{24} \end{array} $$

Remainder is the last entry in the bottom row $ \left(\color{red}{ 24 }\right)$.

Synthetic Division Calculator