Check if $ x+2 $ is a factor of $ x^{3}-11x^{2}+36x-48 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-2&1&-11&36&-48\\& & -2& 26& \color{black}{-124} \\ \hline &\color{blue}{1}&\color{blue}{-13}&\color{blue}{62}&\color{orangered}{-172} \end{array} $$Because the remainder $ \left( \color{red}{ -172 } \right) $ is not zero, we conclude that the $ x+2 $ 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 + 2 = 0 $ ( $ x = \color{blue}{ -2 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&1&-11&36&-48\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-2&\color{orangered}{ 1 }&-11&36&-48\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 1 } = \color{blue}{ -2 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&1&-11&36&-48\\& & \color{blue}{-2} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -11 } + \color{orangered}{ \left( -2 \right) } = \color{orangered}{ -13 } $
$$ \begin{array}{c|rrrr}-2&1&\color{orangered}{ -11 }&36&-48\\& & \color{orangered}{-2} & & \\ \hline &1&\color{orangered}{-13}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -13 \right) } = \color{blue}{ 26 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&1&-11&36&-48\\& & -2& \color{blue}{26} & \\ \hline &1&\color{blue}{-13}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 36 } + \color{orangered}{ 26 } = \color{orangered}{ 62 } $
$$ \begin{array}{c|rrrr}-2&1&-11&\color{orangered}{ 36 }&-48\\& & -2& \color{orangered}{26} & \\ \hline &1&-13&\color{orangered}{62}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 62 } = \color{blue}{ -124 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&1&-11&36&-48\\& & -2& 26& \color{blue}{-124} \\ \hline &1&-13&\color{blue}{62}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -48 } + \color{orangered}{ \left( -124 \right) } = \color{orangered}{ -172 } $
$$ \begin{array}{c|rrrr}-2&1&-11&36&\color{orangered}{ -48 }\\& & -2& 26& \color{orangered}{-124} \\ \hline &\color{blue}{1}&\color{blue}{-13}&\color{blue}{62}&\color{orangered}{-172} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -172 }\right)$.