Check if $ x+2 $ is a factor of $ x^{4}-2x^{3}-25x^{2}+26x+120 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-2&1&-2&-25&26&120\\& & -2& 8& 34& \color{black}{-120} \\ \hline &\color{blue}{1}&\color{blue}{-4}&\color{blue}{-17}&\color{blue}{60}&\color{orangered}{0} \end{array} $$Because the remainder equals zero, we conclude that the $ x+2 $ is a factor of the $ x^{4}-2x^{3}-25x^{2}+26x+120 $.
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.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&1&-2&-25&26&120\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-2&\color{orangered}{ 1 }&-2&-25&26&120\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&1&-2&-25&26&120\\& & \color{blue}{-2} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ \left( -2 \right) } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrrr}-2&1&\color{orangered}{ -2 }&-25&26&120\\& & \color{orangered}{-2} & & & \\ \hline &1&\color{orangered}{-4}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ 8 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&1&-2&-25&26&120\\& & -2& \color{blue}{8} & & \\ \hline &1&\color{blue}{-4}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -25 } + \color{orangered}{ 8 } = \color{orangered}{ -17 } $
$$ \begin{array}{c|rrrrr}-2&1&-2&\color{orangered}{ -25 }&26&120\\& & -2& \color{orangered}{8} & & \\ \hline &1&-4&\color{orangered}{-17}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -17 \right) } = \color{blue}{ 34 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&1&-2&-25&26&120\\& & -2& 8& \color{blue}{34} & \\ \hline &1&-4&\color{blue}{-17}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 26 } + \color{orangered}{ 34 } = \color{orangered}{ 60 } $
$$ \begin{array}{c|rrrrr}-2&1&-2&-25&\color{orangered}{ 26 }&120\\& & -2& 8& \color{orangered}{34} & \\ \hline &1&-4&-17&\color{orangered}{60}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 60 } = \color{blue}{ -120 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&1&-2&-25&26&120\\& & -2& 8& 34& \color{blue}{-120} \\ \hline &1&-4&-17&\color{blue}{60}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 120 } + \color{orangered}{ \left( -120 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}-2&1&-2&-25&26&\color{orangered}{ 120 }\\& & -2& 8& 34& \color{orangered}{-120} \\ \hline &\color{blue}{1}&\color{blue}{-4}&\color{blue}{-17}&\color{blue}{60}&\color{orangered}{0} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 0 }\right)$.