Check if $ x-3 $ is a factor of $ 5x^{3}+25x^{2}-60x-180 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}3&5&25&-60&-180\\& & 15& 120& \color{black}{180} \\ \hline &\color{blue}{5}&\color{blue}{40}&\color{blue}{60}&\color{orangered}{0} \end{array} $$Because the remainder equals zero, we conclude that the $ x-3 $ is a factor of the $ 5x^{3}+25x^{2}-60x-180 $.
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 -3 = 0 $ ( $ x = \color{blue}{ 3 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{3}&5&25&-60&-180\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}3&\color{orangered}{ 5 }&25&-60&-180\\& & & & \\ \hline &\color{orangered}{5}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 5 } = \color{blue}{ 15 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&5&25&-60&-180\\& & \color{blue}{15} & & \\ \hline &\color{blue}{5}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 25 } + \color{orangered}{ 15 } = \color{orangered}{ 40 } $
$$ \begin{array}{c|rrrr}3&5&\color{orangered}{ 25 }&-60&-180\\& & \color{orangered}{15} & & \\ \hline &5&\color{orangered}{40}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 40 } = \color{blue}{ 120 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&5&25&-60&-180\\& & 15& \color{blue}{120} & \\ \hline &5&\color{blue}{40}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -60 } + \color{orangered}{ 120 } = \color{orangered}{ 60 } $
$$ \begin{array}{c|rrrr}3&5&25&\color{orangered}{ -60 }&-180\\& & 15& \color{orangered}{120} & \\ \hline &5&40&\color{orangered}{60}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 60 } = \color{blue}{ 180 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&5&25&-60&-180\\& & 15& 120& \color{blue}{180} \\ \hline &5&40&\color{blue}{60}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -180 } + \color{orangered}{ 180 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}3&5&25&-60&\color{orangered}{ -180 }\\& & 15& 120& \color{orangered}{180} \\ \hline &\color{blue}{5}&\color{blue}{40}&\color{blue}{60}&\color{orangered}{0} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 0 }\right)$.