Divide using synthetic division $$ ( 4x^{3}-19x^{2}+20x+3 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}3&4&-19&20&3\\& & 12& -21& \color{black}{-3} \\ \hline &\color{blue}{4}&\color{blue}{-7}&\color{blue}{-1}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 4x^{3}-19x^{2}+20x+3 }{ x-3 } = \color{blue}{4x^{2}-7x-1} $$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.
$$ \begin{array}{c|rrrr}\color{blue}{3}&4&-19&20&3\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}3&\color{orangered}{ 4 }&-19&20&3\\& & & & \\ \hline &\color{orangered}{4}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&4&-19&20&3\\& & \color{blue}{12} & & \\ \hline &\color{blue}{4}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -19 } + \color{orangered}{ 12 } = \color{orangered}{ -7 } $
$$ \begin{array}{c|rrrr}3&4&\color{orangered}{ -19 }&20&3\\& & \color{orangered}{12} & & \\ \hline &4&\color{orangered}{-7}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -7 \right) } = \color{blue}{ -21 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&4&-19&20&3\\& & 12& \color{blue}{-21} & \\ \hline &4&\color{blue}{-7}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 20 } + \color{orangered}{ \left( -21 \right) } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrr}3&4&-19&\color{orangered}{ 20 }&3\\& & 12& \color{orangered}{-21} & \\ \hline &4&-7&\color{orangered}{-1}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ -3 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&4&-19&20&3\\& & 12& -21& \color{blue}{-3} \\ \hline &4&-7&\color{blue}{-1}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}3&4&-19&20&\color{orangered}{ 3 }\\& & 12& -21& \color{orangered}{-3} \\ \hline &\color{blue}{4}&\color{blue}{-7}&\color{blue}{-1}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{2}-7x-1 } $ with a remainder of $ \color{red}{ 0 } $.