Divide using synthetic division $$ ( x^{4}+6x^{3}-14x^{2}+5x-18 ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}2&1&6&-14&5&-18\\& & 2& 16& 4& \color{black}{18} \\ \hline &\color{blue}{1}&\color{blue}{8}&\color{blue}{2}&\color{blue}{9}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ x^{4}+6x^{3}-14x^{2}+5x-18 }{ x-2 } = \color{blue}{x^{3}+8x^{2}+2x+9} $$Explanation
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.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&6&-14&5&-18\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}2&\color{orangered}{ 1 }&6&-14&5&-18\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&6&-14&5&-18\\& & \color{blue}{2} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ 2 } = \color{orangered}{ 8 } $
$$ \begin{array}{c|rrrrr}2&1&\color{orangered}{ 6 }&-14&5&-18\\& & \color{orangered}{2} & & & \\ \hline &1&\color{orangered}{8}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 8 } = \color{blue}{ 16 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&6&-14&5&-18\\& & 2& \color{blue}{16} & & \\ \hline &1&\color{blue}{8}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -14 } + \color{orangered}{ 16 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrr}2&1&6&\color{orangered}{ -14 }&5&-18\\& & 2& \color{orangered}{16} & & \\ \hline &1&8&\color{orangered}{2}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 2 } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&6&-14&5&-18\\& & 2& 16& \color{blue}{4} & \\ \hline &1&8&\color{blue}{2}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ 4 } = \color{orangered}{ 9 } $
$$ \begin{array}{c|rrrrr}2&1&6&-14&\color{orangered}{ 5 }&-18\\& & 2& 16& \color{orangered}{4} & \\ \hline &1&8&2&\color{orangered}{9}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 9 } = \color{blue}{ 18 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&6&-14&5&-18\\& & 2& 16& 4& \color{blue}{18} \\ \hline &1&8&2&\color{blue}{9}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -18 } + \color{orangered}{ 18 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}2&1&6&-14&5&\color{orangered}{ -18 }\\& & 2& 16& 4& \color{orangered}{18} \\ \hline &\color{blue}{1}&\color{blue}{8}&\color{blue}{2}&\color{blue}{9}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}+8x^{2}+2x+9 } $ with a remainder of $ \color{red}{ 0 } $.