Check if $ x-5 $ is a factor of $ 3x^{4}-15x^{3}-7x+45 $.

The synthetic division table is:

$$ \begin{array}{c|rrrrr}5&3&-15&0&-7&45\\& & 15& 0& 0& \color{black}{-35} \\ \hline &\color{blue}{3}&\color{blue}{0}&\color{blue}{0}&\color{blue}{-7}&\color{orangered}{10} \end{array} $$

Because the remainder $ \left( \color{red}{ 10 } \right) $ is not zero, we conclude that the $ x-5 $ is not a factor of $ 3x^{4}-15x^{3}-7x+45$.

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.

$$ \begin{array}{c|rrrrr}\color{blue}{5}&3&-15&0&-7&45\\& & & & & \\ \hline &&&&& \end{array} $$

Step 1 : Bring down the leading coefficient to the bottom row.

$$ \begin{array}{c|rrrrr}5&\color{orangered}{ 3 }&-15&0&-7&45\\& & & & & \\ \hline &\color{orangered}{3}&&&& \end{array} $$

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 } $.

$$ \begin{array}{c|rrrrr}\color{blue}{5}&3&-15&0&-7&45\\& & \color{blue}{15} & & & \\ \hline &\color{blue}{3}&&&& \end{array} $$

Step 3 : Add down last column: $ \color{orangered}{ -15 } + \color{orangered}{ 15 } = \color{orangered}{ 0 } $

$$ \begin{array}{c|rrrrr}5&3&\color{orangered}{ -15 }&0&-7&45\\& & \color{orangered}{15} & & & \\ \hline &3&\color{orangered}{0}&&& \end{array} $$

Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.

$$ \begin{array}{c|rrrrr}\color{blue}{5}&3&-15&0&-7&45\\& & 15& \color{blue}{0} & & \\ \hline &3&\color{blue}{0}&&& \end{array} $$

Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 0 } = \color{orangered}{ 0 } $

$$ \begin{array}{c|rrrrr}5&3&-15&\color{orangered}{ 0 }&-7&45\\& & 15& \color{orangered}{0} & & \\ \hline &3&0&\color{orangered}{0}&& \end{array} $$

Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.

$$ \begin{array}{c|rrrrr}\color{blue}{5}&3&-15&0&-7&45\\& & 15& 0& \color{blue}{0} & \\ \hline &3&0&\color{blue}{0}&& \end{array} $$

Step 7 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 0 } = \color{orangered}{ -7 } $

$$ \begin{array}{c|rrrrr}5&3&-15&0&\color{orangered}{ -7 }&45\\& & 15& 0& \color{orangered}{0} & \\ \hline &3&0&0&\color{orangered}{-7}& \end{array} $$

Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ \left( -7 \right) } = \color{blue}{ -35 } $.

$$ \begin{array}{c|rrrrr}\color{blue}{5}&3&-15&0&-7&45\\& & 15& 0& 0& \color{blue}{-35} \\ \hline &3&0&0&\color{blue}{-7}& \end{array} $$

Step 9 : Add down last column: $ \color{orangered}{ 45 } + \color{orangered}{ \left( -35 \right) } = \color{orangered}{ 10 } $

$$ \begin{array}{c|rrrrr}5&3&-15&0&-7&\color{orangered}{ 45 }\\& & 15& 0& 0& \color{orangered}{-35} \\ \hline &\color{blue}{3}&\color{blue}{0}&\color{blue}{0}&\color{blue}{-7}&\color{orangered}{10} \end{array} $$

Remainder is the last entry in the bottom row $ \left(\color{red}{ 10 }\right)$.

Synthetic Division Calculator