Divide using synthetic division $$ ( 3x^{3}-11x^{2}-20x+3 ) \div ( x-5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}5&3&-11&-20&3\\& & 15& 20& \color{black}{0} \\ \hline &\color{blue}{3}&\color{blue}{4}&\color{blue}{0}&\color{orangered}{3} \end{array} $$The solution is:
$$ \frac{ 3x^{3}-11x^{2}-20x+3 }{ x-5 } = \color{blue}{3x^{2}+4x} ~+~ \frac{ \color{red}{ 3 } }{ x-5 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -5 = 0 $ ( $ x = \color{blue}{ 5 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{5}&3&-11&-20&3\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}5&\color{orangered}{ 3 }&-11&-20&3\\& & & & \\ \hline &\color{orangered}{3}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 3 } = \color{blue}{ 15 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&3&-11&-20&3\\& & \color{blue}{15} & & \\ \hline &\color{blue}{3}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -11 } + \color{orangered}{ 15 } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrr}5&3&\color{orangered}{ -11 }&-20&3\\& & \color{orangered}{15} & & \\ \hline &3&\color{orangered}{4}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 4 } = \color{blue}{ 20 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&3&-11&-20&3\\& & 15& \color{blue}{20} & \\ \hline &3&\color{blue}{4}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -20 } + \color{orangered}{ 20 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}5&3&-11&\color{orangered}{ -20 }&3\\& & 15& \color{orangered}{20} & \\ \hline &3&4&\color{orangered}{0}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&3&-11&-20&3\\& & 15& 20& \color{blue}{0} \\ \hline &3&4&\color{blue}{0}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 0 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrr}5&3&-11&-20&\color{orangered}{ 3 }\\& & 15& 20& \color{orangered}{0} \\ \hline &\color{blue}{3}&\color{blue}{4}&\color{blue}{0}&\color{orangered}{3} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{2}+4x } $ with a remainder of $ \color{red}{ 3 } $.