Check if $ x-5 $ is a factor of $ 2x^{3}-11x^{2}+13x-44 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}5&2&-11&13&-44\\& & 10& -5& \color{black}{40} \\ \hline &\color{blue}{2}&\color{blue}{-1}&\color{blue}{8}&\color{orangered}{-4} \end{array} $$Because the remainder $ \left( \color{red}{ -4 } \right) $ is not zero, we conclude that the $ x-5 $ is not a factor of $ 2x^{3}-11x^{2}+13x-44$.
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 -5 = 0 $ ( $ x = \color{blue}{ 5 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{5}&2&-11&13&-44\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}5&\color{orangered}{ 2 }&-11&13&-44\\& & & & \\ \hline &\color{orangered}{2}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 2 } = \color{blue}{ 10 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&2&-11&13&-44\\& & \color{blue}{10} & & \\ \hline &\color{blue}{2}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -11 } + \color{orangered}{ 10 } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrr}5&2&\color{orangered}{ -11 }&13&-44\\& & \color{orangered}{10} & & \\ \hline &2&\color{orangered}{-1}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ -5 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&2&-11&13&-44\\& & 10& \color{blue}{-5} & \\ \hline &2&\color{blue}{-1}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 13 } + \color{orangered}{ \left( -5 \right) } = \color{orangered}{ 8 } $
$$ \begin{array}{c|rrrr}5&2&-11&\color{orangered}{ 13 }&-44\\& & 10& \color{orangered}{-5} & \\ \hline &2&-1&\color{orangered}{8}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 8 } = \color{blue}{ 40 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&2&-11&13&-44\\& & 10& -5& \color{blue}{40} \\ \hline &2&-1&\color{blue}{8}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -44 } + \color{orangered}{ 40 } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrr}5&2&-11&13&\color{orangered}{ -44 }\\& & 10& -5& \color{orangered}{40} \\ \hline &\color{blue}{2}&\color{blue}{-1}&\color{blue}{8}&\color{orangered}{-4} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -4 }\right)$.