Check if $ x+12 $ is a factor of $ 5x^{3}-2x+1 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-12&5&0&-2&1\\& & -60& 720& \color{black}{-8616} \\ \hline &\color{blue}{5}&\color{blue}{-60}&\color{blue}{718}&\color{orangered}{-8615} \end{array} $$Because the remainder $ \left( \color{red}{ -8615 } \right) $ is not zero, we conclude that the $ x+12 $ is not a factor of $ 5x^{3}-2x+1$.
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.
$$ \begin{array}{c|rrrr}\color{blue}{-12}&5&0&-2&1\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-12&\color{orangered}{ 5 }&0&-2&1\\& & & & \\ \hline &\color{orangered}{5}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -12 } \cdot \color{blue}{ 5 } = \color{blue}{ -60 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-12}&5&0&-2&1\\& & \color{blue}{-60} & & \\ \hline &\color{blue}{5}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -60 \right) } = \color{orangered}{ -60 } $
$$ \begin{array}{c|rrrr}-12&5&\color{orangered}{ 0 }&-2&1\\& & \color{orangered}{-60} & & \\ \hline &5&\color{orangered}{-60}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -12 } \cdot \color{blue}{ \left( -60 \right) } = \color{blue}{ 720 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-12}&5&0&-2&1\\& & -60& \color{blue}{720} & \\ \hline &5&\color{blue}{-60}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ 720 } = \color{orangered}{ 718 } $
$$ \begin{array}{c|rrrr}-12&5&0&\color{orangered}{ -2 }&1\\& & -60& \color{orangered}{720} & \\ \hline &5&-60&\color{orangered}{718}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -12 } \cdot \color{blue}{ 718 } = \color{blue}{ -8616 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-12}&5&0&-2&1\\& & -60& 720& \color{blue}{-8616} \\ \hline &5&-60&\color{blue}{718}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ \left( -8616 \right) } = \color{orangered}{ -8615 } $
$$ \begin{array}{c|rrrr}-12&5&0&-2&\color{orangered}{ 1 }\\& & -60& 720& \color{orangered}{-8616} \\ \hline &\color{blue}{5}&\color{blue}{-60}&\color{blue}{718}&\color{orangered}{-8615} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -8615 }\right)$.