Check if $ x+3 $ is a factor of $ 2x^{3}-x^{2}+2x+3 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-3&2&-1&2&3\\& & -6& 21& \color{black}{-69} \\ \hline &\color{blue}{2}&\color{blue}{-7}&\color{blue}{23}&\color{orangered}{-66} \end{array} $$Because the remainder $ \left( \color{red}{ -66 } \right) $ is not zero, we conclude that the $ x+3 $ is not a factor of $ 2x^{3}-x^{2}+2x+3$.
Explanation
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.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&2&-1&2&3\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-3&\color{orangered}{ 2 }&-1&2&3\\& & & & \\ \hline &\color{orangered}{2}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&2&-1&2&3\\& & \color{blue}{-6} & & \\ \hline &\color{blue}{2}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ -7 } $
$$ \begin{array}{c|rrrr}-3&2&\color{orangered}{ -1 }&2&3\\& & \color{orangered}{-6} & & \\ \hline &2&\color{orangered}{-7}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -7 \right) } = \color{blue}{ 21 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&2&-1&2&3\\& & -6& \color{blue}{21} & \\ \hline &2&\color{blue}{-7}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 21 } = \color{orangered}{ 23 } $
$$ \begin{array}{c|rrrr}-3&2&-1&\color{orangered}{ 2 }&3\\& & -6& \color{orangered}{21} & \\ \hline &2&-7&\color{orangered}{23}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 23 } = \color{blue}{ -69 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&2&-1&2&3\\& & -6& 21& \color{blue}{-69} \\ \hline &2&-7&\color{blue}{23}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ \left( -69 \right) } = \color{orangered}{ -66 } $
$$ \begin{array}{c|rrrr}-3&2&-1&2&\color{orangered}{ 3 }\\& & -6& 21& \color{orangered}{-69} \\ \hline &\color{blue}{2}&\color{blue}{-7}&\color{blue}{23}&\color{orangered}{-66} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -66 }\right)$.