Check if $ 2x+5 $ is a factor of $ 2x^{3}+9x^{2}+12x+5 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-5&2&9&12&5\\& & -10& 5& \color{black}{-85} \\ \hline &\color{blue}{2}&\color{blue}{-1}&\color{blue}{17}&\color{orangered}{-80} \end{array} $$Because the remainder $ \left( \color{red}{ -80 } \right) $ is not zero, we conclude that the $ x+5 $ is not a factor of $ 2x^{3}+9x^{2}+12x+5$.
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 + 5 = 0 $ ( $ x = \color{blue}{ -5 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&2&9&12&5\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-5&\color{orangered}{ 2 }&9&12&5\\& & & & \\ \hline &\color{orangered}{2}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 2 } = \color{blue}{ -10 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&2&9&12&5\\& & \color{blue}{-10} & & \\ \hline &\color{blue}{2}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 9 } + \color{orangered}{ \left( -10 \right) } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrr}-5&2&\color{orangered}{ 9 }&12&5\\& & \color{orangered}{-10} & & \\ \hline &2&\color{orangered}{-1}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ 5 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&2&9&12&5\\& & -10& \color{blue}{5} & \\ \hline &2&\color{blue}{-1}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ 5 } = \color{orangered}{ 17 } $
$$ \begin{array}{c|rrrr}-5&2&9&\color{orangered}{ 12 }&5\\& & -10& \color{orangered}{5} & \\ \hline &2&-1&\color{orangered}{17}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 17 } = \color{blue}{ -85 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&2&9&12&5\\& & -10& 5& \color{blue}{-85} \\ \hline &2&-1&\color{blue}{17}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ \left( -85 \right) } = \color{orangered}{ -80 } $
$$ \begin{array}{c|rrrr}-5&2&9&12&\color{orangered}{ 5 }\\& & -10& 5& \color{orangered}{-85} \\ \hline &\color{blue}{2}&\color{blue}{-1}&\color{blue}{17}&\color{orangered}{-80} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -80 }\right)$.