Check if $ x $ is a factor of $ 2x^{3}-5x^{2}-6x+15 $.

The synthetic division table is:

$$ \begin{array}{c|rrrr}0&2&-5&-6&15\\& & 0& 0& \color{black}{0} \\ \hline &\color{blue}{2}&\color{blue}{-5}&\color{blue}{-6}&\color{orangered}{15} \end{array} $$

Because the remainder $ \left( \color{red}{ 15 } \right) $ is not zero, we conclude that the $ x $ is not a factor of $ 2x^{3}-5x^{2}-6x+15$.

First we need to create a synthetic division table.

Step 1 : Write down the coefficients of the dividend into division table.Put the zero at the left.

$$ \begin{array}{c|rrrr}\color{blue}{0}&2&-5&-6&15\\& & & & \\ \hline &&&& \end{array} $$

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

$$ \begin{array}{c|rrrr}0&\color{orangered}{ 2 }&-5&-6&15\\& & & & \\ \hline &\color{orangered}{2}&&& \end{array} $$

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

$$ \begin{array}{c|rrrr}\color{blue}{0}&2&-5&-6&15\\& & \color{blue}{0} & & \\ \hline &\color{blue}{2}&&& \end{array} $$

Step 3 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ 0 } = \color{orangered}{ -5 } $

$$ \begin{array}{c|rrrr}0&2&\color{orangered}{ -5 }&-6&15\\& & \color{orangered}{0} & & \\ \hline &2&\color{orangered}{-5}&& \end{array} $$

Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ 0 } $.

$$ \begin{array}{c|rrrr}\color{blue}{0}&2&-5&-6&15\\& & 0& \color{blue}{0} & \\ \hline &2&\color{blue}{-5}&& \end{array} $$

Step 5 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ 0 } = \color{orangered}{ -6 } $

$$ \begin{array}{c|rrrr}0&2&-5&\color{orangered}{ -6 }&15\\& & 0& \color{orangered}{0} & \\ \hline &2&-5&\color{orangered}{-6}& \end{array} $$

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

$$ \begin{array}{c|rrrr}\color{blue}{0}&2&-5&-6&15\\& & 0& 0& \color{blue}{0} \\ \hline &2&-5&\color{blue}{-6}& \end{array} $$

Step 7 : Add down last column: $ \color{orangered}{ 15 } + \color{orangered}{ 0 } = \color{orangered}{ 15 } $

$$ \begin{array}{c|rrrr}0&2&-5&-6&\color{orangered}{ 15 }\\& & 0& 0& \color{orangered}{0} \\ \hline &\color{blue}{2}&\color{blue}{-5}&\color{blue}{-6}&\color{orangered}{15} \end{array} $$

Remainder is the last entry in the bottom row $ \left(\color{red}{ 15 }\right)$.

Synthetic Division Calculator