Check if $ x+3 $ is a factor of $ x^{3}-2x^{2}+x-3 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-3&1&-2&1&-3\\& & -3& 15& \color{black}{-48} \\ \hline &\color{blue}{1}&\color{blue}{-5}&\color{blue}{16}&\color{orangered}{-51} \end{array} $$Because the remainder $ \left( \color{red}{ -51 } \right) $ is not zero, we conclude that the $ x+3 $ is not a factor of $ x^{3}-2x^{2}+x-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.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&1&-2&1&-3\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-3&\color{orangered}{ 1 }&-2&1&-3\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 1 } = \color{blue}{ -3 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&1&-2&1&-3\\& & \color{blue}{-3} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrr}-3&1&\color{orangered}{ -2 }&1&-3\\& & \color{orangered}{-3} & & \\ \hline &1&\color{orangered}{-5}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ 15 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&1&-2&1&-3\\& & -3& \color{blue}{15} & \\ \hline &1&\color{blue}{-5}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 15 } = \color{orangered}{ 16 } $
$$ \begin{array}{c|rrrr}-3&1&-2&\color{orangered}{ 1 }&-3\\& & -3& \color{orangered}{15} & \\ \hline &1&-5&\color{orangered}{16}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 16 } = \color{blue}{ -48 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&1&-2&1&-3\\& & -3& 15& \color{blue}{-48} \\ \hline &1&-5&\color{blue}{16}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ \left( -48 \right) } = \color{orangered}{ -51 } $
$$ \begin{array}{c|rrrr}-3&1&-2&1&\color{orangered}{ -3 }\\& & -3& 15& \color{orangered}{-48} \\ \hline &\color{blue}{1}&\color{blue}{-5}&\color{blue}{16}&\color{orangered}{-51} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -51 }\right)$.