Divide using synthetic division $$ ( 3x^{3}-16x^{2}-9x+3 ) \div ( x-1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}1&3&-16&-9&3\\& & 3& -13& \color{black}{-22} \\ \hline &\color{blue}{3}&\color{blue}{-13}&\color{blue}{-22}&\color{orangered}{-19} \end{array} $$The solution is:
$$ \frac{ 3x^{3}-16x^{2}-9x+3 }{ x-1 } = \color{blue}{3x^{2}-13x-22} \color{red}{~-~} \frac{ \color{red}{ 19 } }{ x-1 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{1}&3&-16&-9&3\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}1&\color{orangered}{ 3 }&-16&-9&3\\& & & & \\ \hline &\color{orangered}{3}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 3 } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&3&-16&-9&3\\& & \color{blue}{3} & & \\ \hline &\color{blue}{3}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -16 } + \color{orangered}{ 3 } = \color{orangered}{ -13 } $
$$ \begin{array}{c|rrrr}1&3&\color{orangered}{ -16 }&-9&3\\& & \color{orangered}{3} & & \\ \hline &3&\color{orangered}{-13}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -13 \right) } = \color{blue}{ -13 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&3&-16&-9&3\\& & 3& \color{blue}{-13} & \\ \hline &3&\color{blue}{-13}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -9 } + \color{orangered}{ \left( -13 \right) } = \color{orangered}{ -22 } $
$$ \begin{array}{c|rrrr}1&3&-16&\color{orangered}{ -9 }&3\\& & 3& \color{orangered}{-13} & \\ \hline &3&-13&\color{orangered}{-22}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -22 \right) } = \color{blue}{ -22 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&3&-16&-9&3\\& & 3& -13& \color{blue}{-22} \\ \hline &3&-13&\color{blue}{-22}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ \left( -22 \right) } = \color{orangered}{ -19 } $
$$ \begin{array}{c|rrrr}1&3&-16&-9&\color{orangered}{ 3 }\\& & 3& -13& \color{orangered}{-22} \\ \hline &\color{blue}{3}&\color{blue}{-13}&\color{blue}{-22}&\color{orangered}{-19} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{2}-13x-22 } $ with a remainder of $ \color{red}{ -19 } $.