Check if $ x+6 $ is a factor of $ x^{3}-2x^{2}-23x-60 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-6&1&-2&-23&-60\\& & -6& 48& \color{black}{-150} \\ \hline &\color{blue}{1}&\color{blue}{-8}&\color{blue}{25}&\color{orangered}{-210} \end{array} $$Because the remainder $ \left( \color{red}{ -210 } \right) $ is not zero, we conclude that the $ x+6 $ 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 + 6 = 0 $ ( $ x = \color{blue}{ -6 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-6}&1&-2&-23&-60\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-6&\color{orangered}{ 1 }&-2&-23&-60\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -6 } \cdot \color{blue}{ 1 } = \color{blue}{ -6 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-6}&1&-2&-23&-60\\& & \color{blue}{-6} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrr}-6&1&\color{orangered}{ -2 }&-23&-60\\& & \color{orangered}{-6} & & \\ \hline &1&\color{orangered}{-8}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -6 } \cdot \color{blue}{ \left( -8 \right) } = \color{blue}{ 48 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-6}&1&-2&-23&-60\\& & -6& \color{blue}{48} & \\ \hline &1&\color{blue}{-8}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -23 } + \color{orangered}{ 48 } = \color{orangered}{ 25 } $
$$ \begin{array}{c|rrrr}-6&1&-2&\color{orangered}{ -23 }&-60\\& & -6& \color{orangered}{48} & \\ \hline &1&-8&\color{orangered}{25}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -6 } \cdot \color{blue}{ 25 } = \color{blue}{ -150 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-6}&1&-2&-23&-60\\& & -6& 48& \color{blue}{-150} \\ \hline &1&-8&\color{blue}{25}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -60 } + \color{orangered}{ \left( -150 \right) } = \color{orangered}{ -210 } $
$$ \begin{array}{c|rrrr}-6&1&-2&-23&\color{orangered}{ -60 }\\& & -6& 48& \color{orangered}{-150} \\ \hline &\color{blue}{1}&\color{blue}{-8}&\color{blue}{25}&\color{orangered}{-210} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -210 }\right)$.