Check if $ x+12 $ is a factor of $ x^{3}-2x^{2}-23x-60 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-12&1&-2&-23&-60\\& & -12& 168& \color{black}{-1740} \\ \hline &\color{blue}{1}&\color{blue}{-14}&\color{blue}{145}&\color{orangered}{-1800} \end{array} $$Because the remainder $ \left( \color{red}{ -1800 } \right) $ is not zero, we conclude that the $ x+12 $ 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 + 12 = 0 $ ( $ x = \color{blue}{ -12 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-12}&1&-2&-23&-60\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-12&\color{orangered}{ 1 }&-2&-23&-60\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -12 } \cdot \color{blue}{ 1 } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-12}&1&-2&-23&-60\\& & \color{blue}{-12} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ -14 } $
$$ \begin{array}{c|rrrr}-12&1&\color{orangered}{ -2 }&-23&-60\\& & \color{orangered}{-12} & & \\ \hline &1&\color{orangered}{-14}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -12 } \cdot \color{blue}{ \left( -14 \right) } = \color{blue}{ 168 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-12}&1&-2&-23&-60\\& & -12& \color{blue}{168} & \\ \hline &1&\color{blue}{-14}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -23 } + \color{orangered}{ 168 } = \color{orangered}{ 145 } $
$$ \begin{array}{c|rrrr}-12&1&-2&\color{orangered}{ -23 }&-60\\& & -12& \color{orangered}{168} & \\ \hline &1&-14&\color{orangered}{145}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -12 } \cdot \color{blue}{ 145 } = \color{blue}{ -1740 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-12}&1&-2&-23&-60\\& & -12& 168& \color{blue}{-1740} \\ \hline &1&-14&\color{blue}{145}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -60 } + \color{orangered}{ \left( -1740 \right) } = \color{orangered}{ -1800 } $
$$ \begin{array}{c|rrrr}-12&1&-2&-23&\color{orangered}{ -60 }\\& & -12& 168& \color{orangered}{-1740} \\ \hline &\color{blue}{1}&\color{blue}{-14}&\color{blue}{145}&\color{orangered}{-1800} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -1800 }\right)$.