Divide using synthetic division $$ ( 12x^{4}-56x^{3}+59x^{2}+9x-18 ) \div ( 2x+1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-1&12&-56&59&9&-18\\& & -12& 68& -127& \color{black}{118} \\ \hline &\color{blue}{12}&\color{blue}{-68}&\color{blue}{127}&\color{blue}{-118}&\color{orangered}{100} \end{array} $$The solution is:
$$ \frac{ 12x^{4}-56x^{3}+59x^{2}+9x-18 }{ x+1 } = \color{blue}{12x^{3}-68x^{2}+127x-118} ~+~ \frac{ \color{red}{ 100 } }{ x+1 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&12&-56&59&9&-18\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-1&\color{orangered}{ 12 }&-56&59&9&-18\\& & & & & \\ \hline &\color{orangered}{12}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 12 } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&12&-56&59&9&-18\\& & \color{blue}{-12} & & & \\ \hline &\color{blue}{12}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -56 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ -68 } $
$$ \begin{array}{c|rrrrr}-1&12&\color{orangered}{ -56 }&59&9&-18\\& & \color{orangered}{-12} & & & \\ \hline &12&\color{orangered}{-68}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ \left( -68 \right) } = \color{blue}{ 68 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&12&-56&59&9&-18\\& & -12& \color{blue}{68} & & \\ \hline &12&\color{blue}{-68}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 59 } + \color{orangered}{ 68 } = \color{orangered}{ 127 } $
$$ \begin{array}{c|rrrrr}-1&12&-56&\color{orangered}{ 59 }&9&-18\\& & -12& \color{orangered}{68} & & \\ \hline &12&-68&\color{orangered}{127}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 127 } = \color{blue}{ -127 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&12&-56&59&9&-18\\& & -12& 68& \color{blue}{-127} & \\ \hline &12&-68&\color{blue}{127}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 9 } + \color{orangered}{ \left( -127 \right) } = \color{orangered}{ -118 } $
$$ \begin{array}{c|rrrrr}-1&12&-56&59&\color{orangered}{ 9 }&-18\\& & -12& 68& \color{orangered}{-127} & \\ \hline &12&-68&127&\color{orangered}{-118}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ \left( -118 \right) } = \color{blue}{ 118 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&12&-56&59&9&-18\\& & -12& 68& -127& \color{blue}{118} \\ \hline &12&-68&127&\color{blue}{-118}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -18 } + \color{orangered}{ 118 } = \color{orangered}{ 100 } $
$$ \begin{array}{c|rrrrr}-1&12&-56&59&9&\color{orangered}{ -18 }\\& & -12& 68& -127& \color{orangered}{118} \\ \hline &\color{blue}{12}&\color{blue}{-68}&\color{blue}{127}&\color{blue}{-118}&\color{orangered}{100} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 12x^{3}-68x^{2}+127x-118 } $ with a remainder of $ \color{red}{ 100 } $.