Check if $ x-1 $ is a factor of $ x^{4}-x^{3}-13x^{2}+25x-12 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}1&1&-1&-13&25&-12\\& & 1& 0& -13& \color{black}{12} \\ \hline &\color{blue}{1}&\color{blue}{0}&\color{blue}{-13}&\color{blue}{12}&\color{orangered}{0} \end{array} $$Because the remainder equals zero, we conclude that the $ x-1 $ is a factor of the $ x^{4}-x^{3}-13x^{2}+25x-12 $.
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.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&1&-1&-13&25&-12\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}1&\color{orangered}{ 1 }&-1&-13&25&-12\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 1 } = \color{blue}{ 1 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&1&-1&-13&25&-12\\& & \color{blue}{1} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 1 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}1&1&\color{orangered}{ -1 }&-13&25&-12\\& & \color{orangered}{1} & & & \\ \hline &1&\color{orangered}{0}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&1&-1&-13&25&-12\\& & 1& \color{blue}{0} & & \\ \hline &1&\color{blue}{0}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -13 } + \color{orangered}{ 0 } = \color{orangered}{ -13 } $
$$ \begin{array}{c|rrrrr}1&1&-1&\color{orangered}{ -13 }&25&-12\\& & 1& \color{orangered}{0} & & \\ \hline &1&0&\color{orangered}{-13}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -13 \right) } = \color{blue}{ -13 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&1&-1&-13&25&-12\\& & 1& 0& \color{blue}{-13} & \\ \hline &1&0&\color{blue}{-13}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 25 } + \color{orangered}{ \left( -13 \right) } = \color{orangered}{ 12 } $
$$ \begin{array}{c|rrrrr}1&1&-1&-13&\color{orangered}{ 25 }&-12\\& & 1& 0& \color{orangered}{-13} & \\ \hline &1&0&-13&\color{orangered}{12}& \end{array} $$Step 8 : 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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&1&-1&-13&25&-12\\& & 1& 0& -13& \color{blue}{12} \\ \hline &1&0&-13&\color{blue}{12}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -12 } + \color{orangered}{ 12 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}1&1&-1&-13&25&\color{orangered}{ -12 }\\& & 1& 0& -13& \color{orangered}{12} \\ \hline &\color{blue}{1}&\color{blue}{0}&\color{blue}{-13}&\color{blue}{12}&\color{orangered}{0} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 0 }\right)$.