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