Check if $ x-1 $ is a factor of $ x^{4}+4x^{3}-7x^{2}-22x+24 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}1&1&4&-7&-22&24\\& & 1& 5& -2& \color{black}{-24} \\ \hline &\color{blue}{1}&\color{blue}{5}&\color{blue}{-2}&\color{blue}{-24}&\color{orangered}{0} \end{array} $$Because the remainder equals zero, we conclude that the $ x-1 $ is a factor of the $ x^{4}+4x^{3}-7x^{2}-22x+24 $.
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.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&1&4&-7&-22&24\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}1&\color{orangered}{ 1 }&4&-7&-22&24\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&1&4&-7&-22&24\\& & \color{blue}{1} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ 1 } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrrr}1&1&\color{orangered}{ 4 }&-7&-22&24\\& & \color{orangered}{1} & & & \\ \hline &1&\color{orangered}{5}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 5 } = \color{blue}{ 5 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&1&4&-7&-22&24\\& & 1& \color{blue}{5} & & \\ \hline &1&\color{blue}{5}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 5 } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrrr}1&1&4&\color{orangered}{ -7 }&-22&24\\& & 1& \color{orangered}{5} & & \\ \hline &1&5&\color{orangered}{-2}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ -2 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&1&4&-7&-22&24\\& & 1& 5& \color{blue}{-2} & \\ \hline &1&5&\color{blue}{-2}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -22 } + \color{orangered}{ \left( -2 \right) } = \color{orangered}{ -24 } $
$$ \begin{array}{c|rrrrr}1&1&4&-7&\color{orangered}{ -22 }&24\\& & 1& 5& \color{orangered}{-2} & \\ \hline &1&5&-2&\color{orangered}{-24}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -24 \right) } = \color{blue}{ -24 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&1&4&-7&-22&24\\& & 1& 5& -2& \color{blue}{-24} \\ \hline &1&5&-2&\color{blue}{-24}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 24 } + \color{orangered}{ \left( -24 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}1&1&4&-7&-22&\color{orangered}{ 24 }\\& & 1& 5& -2& \color{orangered}{-24} \\ \hline &\color{blue}{1}&\color{blue}{5}&\color{blue}{-2}&\color{blue}{-24}&\color{orangered}{0} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 0 }\right)$.