Check if $ x-2 $ is a factor of $ x^{3}-10x^{2}+7x+18 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}2&1&-10&7&18\\& & 2& -16& \color{black}{-18} \\ \hline &\color{blue}{1}&\color{blue}{-8}&\color{blue}{-9}&\color{orangered}{0} \end{array} $$Because the remainder equals zero, we conclude that the $ x-2 $ is a factor of the $ x^{3}-10x^{2}+7x+18 $.
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 -2 = 0 $ ( $ x = \color{blue}{ 2 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{2}&1&-10&7&18\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}2&\color{orangered}{ 1 }&-10&7&18\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 1 } = \color{blue}{ 2 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&1&-10&7&18\\& & \color{blue}{2} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -10 } + \color{orangered}{ 2 } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrr}2&1&\color{orangered}{ -10 }&7&18\\& & \color{orangered}{2} & & \\ \hline &1&\color{orangered}{-8}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -8 \right) } = \color{blue}{ -16 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&1&-10&7&18\\& & 2& \color{blue}{-16} & \\ \hline &1&\color{blue}{-8}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ \left( -16 \right) } = \color{orangered}{ -9 } $
$$ \begin{array}{c|rrrr}2&1&-10&\color{orangered}{ 7 }&18\\& & 2& \color{orangered}{-16} & \\ \hline &1&-8&\color{orangered}{-9}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -9 \right) } = \color{blue}{ -18 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&1&-10&7&18\\& & 2& -16& \color{blue}{-18} \\ \hline &1&-8&\color{blue}{-9}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 18 } + \color{orangered}{ \left( -18 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}2&1&-10&7&\color{orangered}{ 18 }\\& & 2& -16& \color{orangered}{-18} \\ \hline &\color{blue}{1}&\color{blue}{-8}&\color{blue}{-9}&\color{orangered}{0} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 0 }\right)$.