Divide using synthetic division $$ ( 3x^{3}-19x^{2}+33x-9 ) \div ( x-1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}1&3&-19&33&-9\\& & 3& -16& \color{black}{17} \\ \hline &\color{blue}{3}&\color{blue}{-16}&\color{blue}{17}&\color{orangered}{8} \end{array} $$The solution is:
$$ \frac{ 3x^{3}-19x^{2}+33x-9 }{ x-1 } = \color{blue}{3x^{2}-16x+17} ~+~ \frac{ \color{red}{ 8 } }{ x-1 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{1}&3&-19&33&-9\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}1&\color{orangered}{ 3 }&-19&33&-9\\& & & & \\ \hline &\color{orangered}{3}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 3 } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&3&-19&33&-9\\& & \color{blue}{3} & & \\ \hline &\color{blue}{3}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -19 } + \color{orangered}{ 3 } = \color{orangered}{ -16 } $
$$ \begin{array}{c|rrrr}1&3&\color{orangered}{ -19 }&33&-9\\& & \color{orangered}{3} & & \\ \hline &3&\color{orangered}{-16}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -16 \right) } = \color{blue}{ -16 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&3&-19&33&-9\\& & 3& \color{blue}{-16} & \\ \hline &3&\color{blue}{-16}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 33 } + \color{orangered}{ \left( -16 \right) } = \color{orangered}{ 17 } $
$$ \begin{array}{c|rrrr}1&3&-19&\color{orangered}{ 33 }&-9\\& & 3& \color{orangered}{-16} & \\ \hline &3&-16&\color{orangered}{17}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 17 } = \color{blue}{ 17 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&3&-19&33&-9\\& & 3& -16& \color{blue}{17} \\ \hline &3&-16&\color{blue}{17}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -9 } + \color{orangered}{ 17 } = \color{orangered}{ 8 } $
$$ \begin{array}{c|rrrr}1&3&-19&33&\color{orangered}{ -9 }\\& & 3& -16& \color{orangered}{17} \\ \hline &\color{blue}{3}&\color{blue}{-16}&\color{blue}{17}&\color{orangered}{8} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{2}-16x+17 } $ with a remainder of $ \color{red}{ 8 } $.