Check if $ x-4 $ is a factor of $ -2x^{3}+9x^{2}-4x+7 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}4&-2&9&-4&7\\& & -8& 4& \color{black}{0} \\ \hline &\color{blue}{-2}&\color{blue}{1}&\color{blue}{0}&\color{orangered}{7} \end{array} $$Because the remainder $ \left( \color{red}{ 7 } \right) $ is not zero, we conclude that the $ x-4 $ is not a factor of $ -2x^{3}+9x^{2}-4x+7$.
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 -4 = 0 $ ( $ x = \color{blue}{ 4 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{4}&-2&9&-4&7\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}4&\color{orangered}{ -2 }&9&-4&7\\& & & & \\ \hline &\color{orangered}{-2}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ -8 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&-2&9&-4&7\\& & \color{blue}{-8} & & \\ \hline &\color{blue}{-2}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 9 } + \color{orangered}{ \left( -8 \right) } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrr}4&-2&\color{orangered}{ 9 }&-4&7\\& & \color{orangered}{-8} & & \\ \hline &-2&\color{orangered}{1}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 1 } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&-2&9&-4&7\\& & -8& \color{blue}{4} & \\ \hline &-2&\color{blue}{1}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ 4 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}4&-2&9&\color{orangered}{ -4 }&7\\& & -8& \color{orangered}{4} & \\ \hline &-2&1&\color{orangered}{0}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&-2&9&-4&7\\& & -8& 4& \color{blue}{0} \\ \hline &-2&1&\color{blue}{0}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ 0 } = \color{orangered}{ 7 } $
$$ \begin{array}{c|rrrr}4&-2&9&-4&\color{orangered}{ 7 }\\& & -8& 4& \color{orangered}{0} \\ \hline &\color{blue}{-2}&\color{blue}{1}&\color{blue}{0}&\color{orangered}{7} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 7 }\right)$.