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

The synthetic division table is:

$$ \begin{array}{c|rrrr}2&4&-5&0&-2\\& & 8& 6& \color{black}{12} \\ \hline &\color{blue}{4}&\color{blue}{3}&\color{blue}{6}&\color{orangered}{10} \end{array} $$

Because the remainder $ \left( \color{red}{ 10 } \right) $ is not zero, we conclude that the $ x-2 $ is not a factor of $ 4x^{3}-5x^{2}-2$.

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.

$$ \begin{array}{c|rrrr}\color{blue}{2}&4&-5&0&-2\\& & & & \\ \hline &&&& \end{array} $$

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

$$ \begin{array}{c|rrrr}2&\color{orangered}{ 4 }&-5&0&-2\\& & & & \\ \hline &\color{orangered}{4}&&& \end{array} $$

Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 4 } = \color{blue}{ 8 } $.

$$ \begin{array}{c|rrrr}\color{blue}{2}&4&-5&0&-2\\& & \color{blue}{8} & & \\ \hline &\color{blue}{4}&&& \end{array} $$

Step 3 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ 8 } = \color{orangered}{ 3 } $

$$ \begin{array}{c|rrrr}2&4&\color{orangered}{ -5 }&0&-2\\& & \color{orangered}{8} & & \\ \hline &4&\color{orangered}{3}&& \end{array} $$

Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 3 } = \color{blue}{ 6 } $.

$$ \begin{array}{c|rrrr}\color{blue}{2}&4&-5&0&-2\\& & 8& \color{blue}{6} & \\ \hline &4&\color{blue}{3}&& \end{array} $$

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

$$ \begin{array}{c|rrrr}2&4&-5&\color{orangered}{ 0 }&-2\\& & 8& \color{orangered}{6} & \\ \hline &4&3&\color{orangered}{6}& \end{array} $$

Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 6 } = \color{blue}{ 12 } $.

$$ \begin{array}{c|rrrr}\color{blue}{2}&4&-5&0&-2\\& & 8& 6& \color{blue}{12} \\ \hline &4&3&\color{blue}{6}& \end{array} $$

Step 7 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ 12 } = \color{orangered}{ 10 } $

$$ \begin{array}{c|rrrr}2&4&-5&0&\color{orangered}{ -2 }\\& & 8& 6& \color{orangered}{12} \\ \hline &\color{blue}{4}&\color{blue}{3}&\color{blue}{6}&\color{orangered}{10} \end{array} $$

Remainder is the last entry in the bottom row $ \left(\color{red}{ 10 }\right)$.

Synthetic Division Calculator