Check if $ x-1 $ is a factor of $ 12x^{3}+37x^{2}-4x $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}1&12&37&-4&0\\& & 12& 49& \color{black}{45} \\ \hline &\color{blue}{12}&\color{blue}{49}&\color{blue}{45}&\color{orangered}{45} \end{array} $$Because the remainder $ \left( \color{red}{ 45 } \right) $ is not zero, we conclude that the $ x-1 $ is not a factor of $ 12x^{3}+37x^{2}-4x$.
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 -1 = 0 $ ( $ x = \color{blue}{ 1 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{1}&12&37&-4&0\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}1&\color{orangered}{ 12 }&37&-4&0\\& & & & \\ \hline &\color{orangered}{12}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 12 } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&12&37&-4&0\\& & \color{blue}{12} & & \\ \hline &\color{blue}{12}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 37 } + \color{orangered}{ 12 } = \color{orangered}{ 49 } $
$$ \begin{array}{c|rrrr}1&12&\color{orangered}{ 37 }&-4&0\\& & \color{orangered}{12} & & \\ \hline &12&\color{orangered}{49}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 49 } = \color{blue}{ 49 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&12&37&-4&0\\& & 12& \color{blue}{49} & \\ \hline &12&\color{blue}{49}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ 49 } = \color{orangered}{ 45 } $
$$ \begin{array}{c|rrrr}1&12&37&\color{orangered}{ -4 }&0\\& & 12& \color{orangered}{49} & \\ \hline &12&49&\color{orangered}{45}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 45 } = \color{blue}{ 45 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&12&37&-4&0\\& & 12& 49& \color{blue}{45} \\ \hline &12&49&\color{blue}{45}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 45 } = \color{orangered}{ 45 } $
$$ \begin{array}{c|rrrr}1&12&37&-4&\color{orangered}{ 0 }\\& & 12& 49& \color{orangered}{45} \\ \hline &\color{blue}{12}&\color{blue}{49}&\color{blue}{45}&\color{orangered}{45} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 45 }\right)$.