Divide using synthetic division $$ ( 4x^{3}-3x^{2}+20x-20 ) \div ( 4x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}3&4&-3&20&-20\\& & 12& 27& \color{black}{141} \\ \hline &\color{blue}{4}&\color{blue}{9}&\color{blue}{47}&\color{orangered}{121} \end{array} $$The solution is:
$$ \frac{ 4x^{3}-3x^{2}+20x-20 }{ x-3 } = \color{blue}{4x^{2}+9x+47} ~+~ \frac{ \color{red}{ 121 } }{ x-3 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -3 = 0 $ ( $ x = \color{blue}{ 3 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{3}&4&-3&20&-20\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}3&\color{orangered}{ 4 }&-3&20&-20\\& & & & \\ \hline &\color{orangered}{4}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 4 } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&4&-3&20&-20\\& & \color{blue}{12} & & \\ \hline &\color{blue}{4}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ 12 } = \color{orangered}{ 9 } $
$$ \begin{array}{c|rrrr}3&4&\color{orangered}{ -3 }&20&-20\\& & \color{orangered}{12} & & \\ \hline &4&\color{orangered}{9}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 9 } = \color{blue}{ 27 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&4&-3&20&-20\\& & 12& \color{blue}{27} & \\ \hline &4&\color{blue}{9}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 20 } + \color{orangered}{ 27 } = \color{orangered}{ 47 } $
$$ \begin{array}{c|rrrr}3&4&-3&\color{orangered}{ 20 }&-20\\& & 12& \color{orangered}{27} & \\ \hline &4&9&\color{orangered}{47}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 47 } = \color{blue}{ 141 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&4&-3&20&-20\\& & 12& 27& \color{blue}{141} \\ \hline &4&9&\color{blue}{47}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -20 } + \color{orangered}{ 141 } = \color{orangered}{ 121 } $
$$ \begin{array}{c|rrrr}3&4&-3&20&\color{orangered}{ -20 }\\& & 12& 27& \color{orangered}{141} \\ \hline &\color{blue}{4}&\color{blue}{9}&\color{blue}{47}&\color{orangered}{121} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{2}+9x+47 } $ with a remainder of $ \color{red}{ 121 } $.