Check if $ x-1 $ is a factor of $ 2x^{3}-5x^{2}+x+2 $.

The synthetic division table is:

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

Because the remainder equals zero, we conclude that the $ x-1 $ is a factor of the $ 2x^{3}-5x^{2}+x+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 -1 = 0 $ ( $ x = \color{blue}{ 1 } $ ) at the left.

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

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

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

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

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

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

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

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

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

Step 5 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ -2 } $

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

Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ -2 } $.

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

Step 7 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ \left( -2 \right) } = \color{orangered}{ 0 } $

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

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

Synthetic Division Calculator