Check if $ x-5 $ is a factor of $ x^{3}+2x^{2}-23x-60 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}5&1&2&-23&-60\\& & 5& 35& \color{black}{60} \\ \hline &\color{blue}{1}&\color{blue}{7}&\color{blue}{12}&\color{orangered}{0} \end{array} $$Because the remainder equals zero, we conclude that the $ x-5 $ is a factor of the $ x^{3}+2x^{2}-23x-60 $.
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.
$$ \begin{array}{c|rrrr}\color{blue}{5}&1&2&-23&-60\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}5&\color{orangered}{ 1 }&2&-23&-60\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 1 } = \color{blue}{ 5 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&1&2&-23&-60\\& & \color{blue}{5} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 5 } = \color{orangered}{ 7 } $
$$ \begin{array}{c|rrrr}5&1&\color{orangered}{ 2 }&-23&-60\\& & \color{orangered}{5} & & \\ \hline &1&\color{orangered}{7}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 7 } = \color{blue}{ 35 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&1&2&-23&-60\\& & 5& \color{blue}{35} & \\ \hline &1&\color{blue}{7}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -23 } + \color{orangered}{ 35 } = \color{orangered}{ 12 } $
$$ \begin{array}{c|rrrr}5&1&2&\color{orangered}{ -23 }&-60\\& & 5& \color{orangered}{35} & \\ \hline &1&7&\color{orangered}{12}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 12 } = \color{blue}{ 60 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&1&2&-23&-60\\& & 5& 35& \color{blue}{60} \\ \hline &1&7&\color{blue}{12}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -60 } + \color{orangered}{ 60 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}5&1&2&-23&\color{orangered}{ -60 }\\& & 5& 35& \color{orangered}{60} \\ \hline &\color{blue}{1}&\color{blue}{7}&\color{blue}{12}&\color{orangered}{0} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 0 }\right)$.