Check if $ x-3 $ is a factor of $ x^{3}-10x^{2}+7x+18 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}3&1&-10&7&18\\& & 3& -21& \color{black}{-42} \\ \hline &\color{blue}{1}&\color{blue}{-7}&\color{blue}{-14}&\color{orangered}{-24} \end{array} $$Because the remainder $ \left( \color{red}{ -24 } \right) $ is not zero, we conclude that the $ x-3 $ is not a factor of $ 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 -3 = 0 $ ( $ x = \color{blue}{ 3 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&-10&7&18\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}3&\color{orangered}{ 1 }&-10&7&18\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 1 } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&-10&7&18\\& & \color{blue}{3} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -10 } + \color{orangered}{ 3 } = \color{orangered}{ -7 } $
$$ \begin{array}{c|rrrr}3&1&\color{orangered}{ -10 }&7&18\\& & \color{orangered}{3} & & \\ \hline &1&\color{orangered}{-7}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -7 \right) } = \color{blue}{ -21 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&-10&7&18\\& & 3& \color{blue}{-21} & \\ \hline &1&\color{blue}{-7}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ \left( -21 \right) } = \color{orangered}{ -14 } $
$$ \begin{array}{c|rrrr}3&1&-10&\color{orangered}{ 7 }&18\\& & 3& \color{orangered}{-21} & \\ \hline &1&-7&\color{orangered}{-14}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -14 \right) } = \color{blue}{ -42 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&-10&7&18\\& & 3& -21& \color{blue}{-42} \\ \hline &1&-7&\color{blue}{-14}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 18 } + \color{orangered}{ \left( -42 \right) } = \color{orangered}{ -24 } $
$$ \begin{array}{c|rrrr}3&1&-10&7&\color{orangered}{ 18 }\\& & 3& -21& \color{orangered}{-42} \\ \hline &\color{blue}{1}&\color{blue}{-7}&\color{blue}{-14}&\color{orangered}{-24} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -24 }\right)$.