Check if $ 2x-1 $ is a factor of $ 4x^{4}+12x^{3}+9x^{2}-8x-5 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}1&4&12&9&-8&-5\\& & 4& 16& 25& \color{black}{17} \\ \hline &\color{blue}{4}&\color{blue}{16}&\color{blue}{25}&\color{blue}{17}&\color{orangered}{12} \end{array} $$Because the remainder $ \left( \color{red}{ 12 } \right) $ is not zero, we conclude that the $ x-1 $ is not a factor of $ 4x^{4}+12x^{3}+9x^{2}-8x-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 -1 = 0 $ ( $ x = \color{blue}{ 1 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&4&12&9&-8&-5\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}1&\color{orangered}{ 4 }&12&9&-8&-5\\& & & & & \\ \hline &\color{orangered}{4}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 4 } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&4&12&9&-8&-5\\& & \color{blue}{4} & & & \\ \hline &\color{blue}{4}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ 4 } = \color{orangered}{ 16 } $
$$ \begin{array}{c|rrrrr}1&4&\color{orangered}{ 12 }&9&-8&-5\\& & \color{orangered}{4} & & & \\ \hline &4&\color{orangered}{16}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 16 } = \color{blue}{ 16 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&4&12&9&-8&-5\\& & 4& \color{blue}{16} & & \\ \hline &4&\color{blue}{16}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 9 } + \color{orangered}{ 16 } = \color{orangered}{ 25 } $
$$ \begin{array}{c|rrrrr}1&4&12&\color{orangered}{ 9 }&-8&-5\\& & 4& \color{orangered}{16} & & \\ \hline &4&16&\color{orangered}{25}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 25 } = \color{blue}{ 25 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&4&12&9&-8&-5\\& & 4& 16& \color{blue}{25} & \\ \hline &4&16&\color{blue}{25}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -8 } + \color{orangered}{ 25 } = \color{orangered}{ 17 } $
$$ \begin{array}{c|rrrrr}1&4&12&9&\color{orangered}{ -8 }&-5\\& & 4& 16& \color{orangered}{25} & \\ \hline &4&16&25&\color{orangered}{17}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 17 } = \color{blue}{ 17 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&4&12&9&-8&-5\\& & 4& 16& 25& \color{blue}{17} \\ \hline &4&16&25&\color{blue}{17}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ 17 } = \color{orangered}{ 12 } $
$$ \begin{array}{c|rrrrr}1&4&12&9&-8&\color{orangered}{ -5 }\\& & 4& 16& 25& \color{orangered}{17} \\ \hline &\color{blue}{4}&\color{blue}{16}&\color{blue}{25}&\color{blue}{17}&\color{orangered}{12} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 12 }\right)$.