Divide using synthetic division $$ ( 2x^{4}+7x^{3}-2x^{2}-17x-18 ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}2&2&7&-2&-17&-18\\& & 4& 22& 40& \color{black}{46} \\ \hline &\color{blue}{2}&\color{blue}{11}&\color{blue}{20}&\color{blue}{23}&\color{orangered}{28} \end{array} $$The solution is:
$$ \frac{ 2x^{4}+7x^{3}-2x^{2}-17x-18 }{ x-2 } = \color{blue}{2x^{3}+11x^{2}+20x+23} ~+~ \frac{ \color{red}{ 28 } }{ x-2 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&2&7&-2&-17&-18\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}2&\color{orangered}{ 2 }&7&-2&-17&-18\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 2 } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&2&7&-2&-17&-18\\& & \color{blue}{4} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ 4 } = \color{orangered}{ 11 } $
$$ \begin{array}{c|rrrrr}2&2&\color{orangered}{ 7 }&-2&-17&-18\\& & \color{orangered}{4} & & & \\ \hline &2&\color{orangered}{11}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 11 } = \color{blue}{ 22 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&2&7&-2&-17&-18\\& & 4& \color{blue}{22} & & \\ \hline &2&\color{blue}{11}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ 22 } = \color{orangered}{ 20 } $
$$ \begin{array}{c|rrrrr}2&2&7&\color{orangered}{ -2 }&-17&-18\\& & 4& \color{orangered}{22} & & \\ \hline &2&11&\color{orangered}{20}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 20 } = \color{blue}{ 40 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&2&7&-2&-17&-18\\& & 4& 22& \color{blue}{40} & \\ \hline &2&11&\color{blue}{20}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -17 } + \color{orangered}{ 40 } = \color{orangered}{ 23 } $
$$ \begin{array}{c|rrrrr}2&2&7&-2&\color{orangered}{ -17 }&-18\\& & 4& 22& \color{orangered}{40} & \\ \hline &2&11&20&\color{orangered}{23}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 23 } = \color{blue}{ 46 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&2&7&-2&-17&-18\\& & 4& 22& 40& \color{blue}{46} \\ \hline &2&11&20&\color{blue}{23}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -18 } + \color{orangered}{ 46 } = \color{orangered}{ 28 } $
$$ \begin{array}{c|rrrrr}2&2&7&-2&-17&\color{orangered}{ -18 }\\& & 4& 22& 40& \color{orangered}{46} \\ \hline &\color{blue}{2}&\color{blue}{11}&\color{blue}{20}&\color{blue}{23}&\color{orangered}{28} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{3}+11x^{2}+20x+23 } $ with a remainder of $ \color{red}{ 28 } $.