Divide using synthetic division $$ ( 4x^{3}+12x^{2}-3x-4 ) \div ( 2x+1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-1&4&12&-3&-4\\& & -4& -8& \color{black}{11} \\ \hline &\color{blue}{4}&\color{blue}{8}&\color{blue}{-11}&\color{orangered}{7} \end{array} $$The solution is:
$$ \frac{ 4x^{3}+12x^{2}-3x-4 }{ x+1 } = \color{blue}{4x^{2}+8x-11} ~+~ \frac{ \color{red}{ 7 } }{ x+1 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&4&12&-3&-4\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-1&\color{orangered}{ 4 }&12&-3&-4\\& & & & \\ \hline &\color{orangered}{4}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 4 } = \color{blue}{ -4 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&4&12&-3&-4\\& & \color{blue}{-4} & & \\ \hline &\color{blue}{4}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ 8 } $
$$ \begin{array}{c|rrrr}-1&4&\color{orangered}{ 12 }&-3&-4\\& & \color{orangered}{-4} & & \\ \hline &4&\color{orangered}{8}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 8 } = \color{blue}{ -8 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&4&12&-3&-4\\& & -4& \color{blue}{-8} & \\ \hline &4&\color{blue}{8}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ \left( -8 \right) } = \color{orangered}{ -11 } $
$$ \begin{array}{c|rrrr}-1&4&12&\color{orangered}{ -3 }&-4\\& & -4& \color{orangered}{-8} & \\ \hline &4&8&\color{orangered}{-11}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ \left( -11 \right) } = \color{blue}{ 11 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&4&12&-3&-4\\& & -4& -8& \color{blue}{11} \\ \hline &4&8&\color{blue}{-11}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ 11 } = \color{orangered}{ 7 } $
$$ \begin{array}{c|rrrr}-1&4&12&-3&\color{orangered}{ -4 }\\& & -4& -8& \color{orangered}{11} \\ \hline &\color{blue}{4}&\color{blue}{8}&\color{blue}{-11}&\color{orangered}{7} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{2}+8x-11 } $ with a remainder of $ \color{red}{ 7 } $.