Divide using synthetic division $$ ( 4x^{3}+12x^{2}-3x-4 ) \div ( 2x-1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}1&4&12&-3&-4\\& & 4& 16& \color{black}{13} \\ \hline &\color{blue}{4}&\color{blue}{16}&\color{blue}{13}&\color{orangered}{9} \end{array} $$The solution is:
$$ \frac{ 4x^{3}+12x^{2}-3x-4 }{ x-1 } = \color{blue}{4x^{2}+16x+13} ~+~ \frac{ \color{red}{ 9 } }{ x-1 } $$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}{ 4 } = \color{orangered}{ 16 } $
$$ \begin{array}{c|rrrr}1&4&\color{orangered}{ 12 }&-3&-4\\& & \color{orangered}{4} & & \\ \hline &4&\color{orangered}{16}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 16 } = \color{blue}{ 16 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&4&12&-3&-4\\& & 4& \color{blue}{16} & \\ \hline &4&\color{blue}{16}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ 16 } = \color{orangered}{ 13 } $
$$ \begin{array}{c|rrrr}1&4&12&\color{orangered}{ -3 }&-4\\& & 4& \color{orangered}{16} & \\ \hline &4&16&\color{orangered}{13}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 13 } = \color{blue}{ 13 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&4&12&-3&-4\\& & 4& 16& \color{blue}{13} \\ \hline &4&16&\color{blue}{13}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ 13 } = \color{orangered}{ 9 } $
$$ \begin{array}{c|rrrr}1&4&12&-3&\color{orangered}{ -4 }\\& & 4& 16& \color{orangered}{13} \\ \hline &\color{blue}{4}&\color{blue}{16}&\color{blue}{13}&\color{orangered}{9} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{2}+16x+13 } $ with a remainder of $ \color{red}{ 9 } $.