Check if $ x-3 $ is a factor of $ x^{3}-x^{2}+99x-152 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}3&1&-1&99&-152\\& & 3& 6& \color{black}{315} \\ \hline &\color{blue}{1}&\color{blue}{2}&\color{blue}{105}&\color{orangered}{163} \end{array} $$Because the remainder $ \left( \color{red}{ 163 } \right) $ is not zero, we conclude that the $ x-3 $ is not a factor of $ x^{3}-x^{2}+99x-152$.
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.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&-1&99&-152\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}3&\color{orangered}{ 1 }&-1&99&-152\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&-1&99&-152\\& & \color{blue}{3} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 3 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrr}3&1&\color{orangered}{ -1 }&99&-152\\& & \color{orangered}{3} & & \\ \hline &1&\color{orangered}{2}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 2 } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&-1&99&-152\\& & 3& \color{blue}{6} & \\ \hline &1&\color{blue}{2}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 99 } + \color{orangered}{ 6 } = \color{orangered}{ 105 } $
$$ \begin{array}{c|rrrr}3&1&-1&\color{orangered}{ 99 }&-152\\& & 3& \color{orangered}{6} & \\ \hline &1&2&\color{orangered}{105}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 105 } = \color{blue}{ 315 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&-1&99&-152\\& & 3& 6& \color{blue}{315} \\ \hline &1&2&\color{blue}{105}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -152 } + \color{orangered}{ 315 } = \color{orangered}{ 163 } $
$$ \begin{array}{c|rrrr}3&1&-1&99&\color{orangered}{ -152 }\\& & 3& 6& \color{orangered}{315} \\ \hline &\color{blue}{1}&\color{blue}{2}&\color{blue}{105}&\color{orangered}{163} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 163 }\right)$.