Check if $ x-5 $ is a factor of $ 3x^{4}-14x^{2}-5 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}5&3&0&-14&0&-5\\& & 15& 75& 305& \color{black}{1525} \\ \hline &\color{blue}{3}&\color{blue}{15}&\color{blue}{61}&\color{blue}{305}&\color{orangered}{1520} \end{array} $$Because the remainder $ \left( \color{red}{ 1520 } \right) $ is not zero, we conclude that the $ x-5 $ is not a factor of $ 3x^{4}-14x^{2}-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.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&3&0&-14&0&-5\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}5&\color{orangered}{ 3 }&0&-14&0&-5\\& & & & & \\ \hline &\color{orangered}{3}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 3 } = \color{blue}{ 15 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&3&0&-14&0&-5\\& & \color{blue}{15} & & & \\ \hline &\color{blue}{3}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 15 } = \color{orangered}{ 15 } $
$$ \begin{array}{c|rrrrr}5&3&\color{orangered}{ 0 }&-14&0&-5\\& & \color{orangered}{15} & & & \\ \hline &3&\color{orangered}{15}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 15 } = \color{blue}{ 75 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&3&0&-14&0&-5\\& & 15& \color{blue}{75} & & \\ \hline &3&\color{blue}{15}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -14 } + \color{orangered}{ 75 } = \color{orangered}{ 61 } $
$$ \begin{array}{c|rrrrr}5&3&0&\color{orangered}{ -14 }&0&-5\\& & 15& \color{orangered}{75} & & \\ \hline &3&15&\color{orangered}{61}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 61 } = \color{blue}{ 305 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&3&0&-14&0&-5\\& & 15& 75& \color{blue}{305} & \\ \hline &3&15&\color{blue}{61}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 305 } = \color{orangered}{ 305 } $
$$ \begin{array}{c|rrrrr}5&3&0&-14&\color{orangered}{ 0 }&-5\\& & 15& 75& \color{orangered}{305} & \\ \hline &3&15&61&\color{orangered}{305}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 305 } = \color{blue}{ 1525 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&3&0&-14&0&-5\\& & 15& 75& 305& \color{blue}{1525} \\ \hline &3&15&61&\color{blue}{305}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ 1525 } = \color{orangered}{ 1520 } $
$$ \begin{array}{c|rrrrr}5&3&0&-14&0&\color{orangered}{ -5 }\\& & 15& 75& 305& \color{orangered}{1525} \\ \hline &\color{blue}{3}&\color{blue}{15}&\color{blue}{61}&\color{blue}{305}&\color{orangered}{1520} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 1520 }\right)$.