Check if $ x+10 $ is a factor of $ 2x^{3}+21x^{2}+5x-50 $.
The synthetic division table is:
$$ \begin{array}{c|rrrr}-10&2&21&5&-50\\& & -20& -10& \color{black}{50} \\ \hline &\color{blue}{2}&\color{blue}{1}&\color{blue}{-5}&\color{orangered}{0} \end{array} $$Because the remainder equals zero, we conclude that the $ x+10 $ is a factor of the $ 2x^{3}+21x^{2}+5x-50 $.
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 + 10 = 0 $ ( $ x = \color{blue}{ -10 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-10}&2&21&5&-50\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-10&\color{orangered}{ 2 }&21&5&-50\\& & & & \\ \hline &\color{orangered}{2}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -10 } \cdot \color{blue}{ 2 } = \color{blue}{ -20 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-10}&2&21&5&-50\\& & \color{blue}{-20} & & \\ \hline &\color{blue}{2}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 21 } + \color{orangered}{ \left( -20 \right) } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrr}-10&2&\color{orangered}{ 21 }&5&-50\\& & \color{orangered}{-20} & & \\ \hline &2&\color{orangered}{1}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -10 } \cdot \color{blue}{ 1 } = \color{blue}{ -10 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-10}&2&21&5&-50\\& & -20& \color{blue}{-10} & \\ \hline &2&\color{blue}{1}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ \left( -10 \right) } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrr}-10&2&21&\color{orangered}{ 5 }&-50\\& & -20& \color{orangered}{-10} & \\ \hline &2&1&\color{orangered}{-5}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -10 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ 50 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-10}&2&21&5&-50\\& & -20& -10& \color{blue}{50} \\ \hline &2&1&\color{blue}{-5}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -50 } + \color{orangered}{ 50 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}-10&2&21&5&\color{orangered}{ -50 }\\& & -20& -10& \color{orangered}{50} \\ \hline &\color{blue}{2}&\color{blue}{1}&\color{blue}{-5}&\color{orangered}{0} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 0 }\right)$.