Check if $ x-7 $ is a factor of $ x^{3}-11x^{2}+36x-48 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}7&1&-11&36&-48\\& & 7& -28& \color{black}{56} \\ \hline &\color{blue}{1}&\color{blue}{-4}&\color{blue}{8}&\color{orangered}{8} \end{array} $$Because the remainder $ \left( \color{red}{ 8 } \right) $ is not zero, we conclude that the $ x-7 $ 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 -7 = 0 $ ( $ x = \color{blue}{ 7 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{7}&1&-11&36&-48\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}7&\color{orangered}{ 1 }&-11&36&-48\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 7 } \cdot \color{blue}{ 1 } = \color{blue}{ 7 } $.
$$ \begin{array}{c|rrrr}\color{blue}{7}&1&-11&36&-48\\& & \color{blue}{7} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -11 } + \color{orangered}{ 7 } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrr}7&1&\color{orangered}{ -11 }&36&-48\\& & \color{orangered}{7} & & \\ \hline &1&\color{orangered}{-4}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 7 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ -28 } $.
$$ \begin{array}{c|rrrr}\color{blue}{7}&1&-11&36&-48\\& & 7& \color{blue}{-28} & \\ \hline &1&\color{blue}{-4}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 36 } + \color{orangered}{ \left( -28 \right) } = \color{orangered}{ 8 } $
$$ \begin{array}{c|rrrr}7&1&-11&\color{orangered}{ 36 }&-48\\& & 7& \color{orangered}{-28} & \\ \hline &1&-4&\color{orangered}{8}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 7 } \cdot \color{blue}{ 8 } = \color{blue}{ 56 } $.
$$ \begin{array}{c|rrrr}\color{blue}{7}&1&-11&36&-48\\& & 7& -28& \color{blue}{56} \\ \hline &1&-4&\color{blue}{8}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -48 } + \color{orangered}{ 56 } = \color{orangered}{ 8 } $
$$ \begin{array}{c|rrrr}7&1&-11&36&\color{orangered}{ -48 }\\& & 7& -28& \color{orangered}{56} \\ \hline &\color{blue}{1}&\color{blue}{-4}&\color{blue}{8}&\color{orangered}{8} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 8 }\right)$.