Check if $ x-3 $ is a factor of $ x^{3}-2x^{2}-23x-60 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}3&1&-2&-23&-60\\& & 3& 3& \color{black}{-60} \\ \hline &\color{blue}{1}&\color{blue}{1}&\color{blue}{-20}&\color{orangered}{-120} \end{array} $$Because the remainder $ \left( \color{red}{ -120 } \right) $ is not zero, we conclude that the $ x-3 $ is not a factor of $ x^{3}-2x^{2}-23x-60$.
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.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&-2&-23&-60\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}3&\color{orangered}{ 1 }&-2&-23&-60\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&-2&-23&-60\\& & \color{blue}{3} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ 3 } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrr}3&1&\color{orangered}{ -2 }&-23&-60\\& & \color{orangered}{3} & & \\ \hline &1&\color{orangered}{1}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 1 } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&-2&-23&-60\\& & 3& \color{blue}{3} & \\ \hline &1&\color{blue}{1}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -23 } + \color{orangered}{ 3 } = \color{orangered}{ -20 } $
$$ \begin{array}{c|rrrr}3&1&-2&\color{orangered}{ -23 }&-60\\& & 3& \color{orangered}{3} & \\ \hline &1&1&\color{orangered}{-20}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -20 \right) } = \color{blue}{ -60 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&-2&-23&-60\\& & 3& 3& \color{blue}{-60} \\ \hline &1&1&\color{blue}{-20}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -60 } + \color{orangered}{ \left( -60 \right) } = \color{orangered}{ -120 } $
$$ \begin{array}{c|rrrr}3&1&-2&-23&\color{orangered}{ -60 }\\& & 3& 3& \color{orangered}{-60} \\ \hline &\color{blue}{1}&\color{blue}{1}&\color{blue}{-20}&\color{orangered}{-120} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -120 }\right)$.