Divide using synthetic division $$ ( 2x^{4}-3x^{3}-14x^{2}+33x-18 ) \div ( x+1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-1&2&-3&-14&33&-18\\& & -2& 5& 9& \color{black}{-42} \\ \hline &\color{blue}{2}&\color{blue}{-5}&\color{blue}{-9}&\color{blue}{42}&\color{orangered}{-60} \end{array} $$The solution is:
$$ \frac{ 2x^{4}-3x^{3}-14x^{2}+33x-18 }{ x+1 } = \color{blue}{2x^{3}-5x^{2}-9x+42} \color{red}{~-~} \frac{ \color{red}{ 60 } }{ x+1 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 1 = 0 $ ( $ x = \color{blue}{ -1 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&2&-3&-14&33&-18\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-1&\color{orangered}{ 2 }&-3&-14&33&-18\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 2 } = \color{blue}{ -2 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&2&-3&-14&33&-18\\& & \color{blue}{-2} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ \left( -2 \right) } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrrr}-1&2&\color{orangered}{ -3 }&-14&33&-18\\& & \color{orangered}{-2} & & & \\ \hline &2&\color{orangered}{-5}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ 5 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&2&-3&-14&33&-18\\& & -2& \color{blue}{5} & & \\ \hline &2&\color{blue}{-5}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -14 } + \color{orangered}{ 5 } = \color{orangered}{ -9 } $
$$ \begin{array}{c|rrrrr}-1&2&-3&\color{orangered}{ -14 }&33&-18\\& & -2& \color{orangered}{5} & & \\ \hline &2&-5&\color{orangered}{-9}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ \left( -9 \right) } = \color{blue}{ 9 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&2&-3&-14&33&-18\\& & -2& 5& \color{blue}{9} & \\ \hline &2&-5&\color{blue}{-9}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 33 } + \color{orangered}{ 9 } = \color{orangered}{ 42 } $
$$ \begin{array}{c|rrrrr}-1&2&-3&-14&\color{orangered}{ 33 }&-18\\& & -2& 5& \color{orangered}{9} & \\ \hline &2&-5&-9&\color{orangered}{42}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 42 } = \color{blue}{ -42 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&2&-3&-14&33&-18\\& & -2& 5& 9& \color{blue}{-42} \\ \hline &2&-5&-9&\color{blue}{42}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -18 } + \color{orangered}{ \left( -42 \right) } = \color{orangered}{ -60 } $
$$ \begin{array}{c|rrrrr}-1&2&-3&-14&33&\color{orangered}{ -18 }\\& & -2& 5& 9& \color{orangered}{-42} \\ \hline &\color{blue}{2}&\color{blue}{-5}&\color{blue}{-9}&\color{blue}{42}&\color{orangered}{-60} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{3}-5x^{2}-9x+42 } $ with a remainder of $ \color{red}{ -60 } $.