Divide using synthetic division $$ ( 2x^{4}+7x^{3}-24x^{2}-27x-18 ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}2&2&7&-24&-27&-18\\& & 4& 22& -4& \color{black}{-62} \\ \hline &\color{blue}{2}&\color{blue}{11}&\color{blue}{-2}&\color{blue}{-31}&\color{orangered}{-80} \end{array} $$The solution is:
$$ \frac{ 2x^{4}+7x^{3}-24x^{2}-27x-18 }{ x-2 } = \color{blue}{2x^{3}+11x^{2}-2x-31} \color{red}{~-~} \frac{ \color{red}{ 80 } }{ x-2 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&2&7&-24&-27&-18\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}2&\color{orangered}{ 2 }&7&-24&-27&-18\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&2&7&-24&-27&-18\\& & \color{blue}{4} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ 4 } = \color{orangered}{ 11 } $
$$ \begin{array}{c|rrrrr}2&2&\color{orangered}{ 7 }&-24&-27&-18\\& & \color{orangered}{4} & & & \\ \hline &2&\color{orangered}{11}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&2&7&-24&-27&-18\\& & 4& \color{blue}{22} & & \\ \hline &2&\color{blue}{11}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -24 } + \color{orangered}{ 22 } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrrr}2&2&7&\color{orangered}{ -24 }&-27&-18\\& & 4& \color{orangered}{22} & & \\ \hline &2&11&\color{orangered}{-2}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ -4 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&2&7&-24&-27&-18\\& & 4& 22& \color{blue}{-4} & \\ \hline &2&11&\color{blue}{-2}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -27 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ -31 } $
$$ \begin{array}{c|rrrrr}2&2&7&-24&\color{orangered}{ -27 }&-18\\& & 4& 22& \color{orangered}{-4} & \\ \hline &2&11&-2&\color{orangered}{-31}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -31 \right) } = \color{blue}{ -62 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&2&7&-24&-27&-18\\& & 4& 22& -4& \color{blue}{-62} \\ \hline &2&11&-2&\color{blue}{-31}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -18 } + \color{orangered}{ \left( -62 \right) } = \color{orangered}{ -80 } $
$$ \begin{array}{c|rrrrr}2&2&7&-24&-27&\color{orangered}{ -18 }\\& & 4& 22& -4& \color{orangered}{-62} \\ \hline &\color{blue}{2}&\color{blue}{11}&\color{blue}{-2}&\color{blue}{-31}&\color{orangered}{-80} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{3}+11x^{2}-2x-31 } $ with a remainder of $ \color{red}{ -80 } $.