Check if $ x-8 $ is a factor of $ x^{3}-11x^{2}+36x-48 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}8&1&-11&36&-48\\& & 8& -24& \color{black}{96} \\ \hline &\color{blue}{1}&\color{blue}{-3}&\color{blue}{12}&\color{orangered}{48} \end{array} $$Because the remainder $ \left( \color{red}{ 48 } \right) $ is not zero, we conclude that the $ x-8 $ is not a factor of $ x^{3}-11x^{2}+36x-48$.
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 -8 = 0 $ ( $ x = \color{blue}{ 8 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{8}&1&-11&36&-48\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}8&\color{orangered}{ 1 }&-11&36&-48\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 8 } \cdot \color{blue}{ 1 } = \color{blue}{ 8 } $.
$$ \begin{array}{c|rrrr}\color{blue}{8}&1&-11&36&-48\\& & \color{blue}{8} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -11 } + \color{orangered}{ 8 } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrr}8&1&\color{orangered}{ -11 }&36&-48\\& & \color{orangered}{8} & & \\ \hline &1&\color{orangered}{-3}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 8 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ -24 } $.
$$ \begin{array}{c|rrrr}\color{blue}{8}&1&-11&36&-48\\& & 8& \color{blue}{-24} & \\ \hline &1&\color{blue}{-3}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 36 } + \color{orangered}{ \left( -24 \right) } = \color{orangered}{ 12 } $
$$ \begin{array}{c|rrrr}8&1&-11&\color{orangered}{ 36 }&-48\\& & 8& \color{orangered}{-24} & \\ \hline &1&-3&\color{orangered}{12}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 8 } \cdot \color{blue}{ 12 } = \color{blue}{ 96 } $.
$$ \begin{array}{c|rrrr}\color{blue}{8}&1&-11&36&-48\\& & 8& -24& \color{blue}{96} \\ \hline &1&-3&\color{blue}{12}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -48 } + \color{orangered}{ 96 } = \color{orangered}{ 48 } $
$$ \begin{array}{c|rrrr}8&1&-11&36&\color{orangered}{ -48 }\\& & 8& -24& \color{orangered}{96} \\ \hline &\color{blue}{1}&\color{blue}{-3}&\color{blue}{12}&\color{orangered}{48} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 48 }\right)$.