Check if $ x-3 $ is a factor of $ 2x^{3}-5x^{2}+x-12 $.

The synthetic division table is:

$$ \begin{array}{c|rrrr}3&2&-5&1&-12\\& & 6& 3& \color{black}{12} \\ \hline &\color{blue}{2}&\color{blue}{1}&\color{blue}{4}&\color{orangered}{0} \end{array} $$

Because the remainder equals zero, we conclude that the $ x-3 $ is a factor of the $ 2x^{3}-5x^{2}+x-12 $.

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.

$$ \begin{array}{c|rrrr}\color{blue}{3}&2&-5&1&-12\\& & & & \\ \hline &&&& \end{array} $$

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

$$ \begin{array}{c|rrrr}3&\color{orangered}{ 2 }&-5&1&-12\\& & & & \\ \hline &\color{orangered}{2}&&& \end{array} $$

Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 2 } = \color{blue}{ 6 } $.

$$ \begin{array}{c|rrrr}\color{blue}{3}&2&-5&1&-12\\& & \color{blue}{6} & & \\ \hline &\color{blue}{2}&&& \end{array} $$

Step 3 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ 6 } = \color{orangered}{ 1 } $

$$ \begin{array}{c|rrrr}3&2&\color{orangered}{ -5 }&1&-12\\& & \color{orangered}{6} & & \\ \hline &2&\color{orangered}{1}&& \end{array} $$

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

$$ \begin{array}{c|rrrr}\color{blue}{3}&2&-5&1&-12\\& & 6& \color{blue}{3} & \\ \hline &2&\color{blue}{1}&& \end{array} $$

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

$$ \begin{array}{c|rrrr}3&2&-5&\color{orangered}{ 1 }&-12\\& & 6& \color{orangered}{3} & \\ \hline &2&1&\color{orangered}{4}& \end{array} $$

Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 4 } = \color{blue}{ 12 } $.

$$ \begin{array}{c|rrrr}\color{blue}{3}&2&-5&1&-12\\& & 6& 3& \color{blue}{12} \\ \hline &2&1&\color{blue}{4}& \end{array} $$

Step 7 : Add down last column: $ \color{orangered}{ -12 } + \color{orangered}{ 12 } = \color{orangered}{ 0 } $

$$ \begin{array}{c|rrrr}3&2&-5&1&\color{orangered}{ -12 }\\& & 6& 3& \color{orangered}{12} \\ \hline &\color{blue}{2}&\color{blue}{1}&\color{blue}{4}&\color{orangered}{0} \end{array} $$

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

Synthetic Division Calculator