Divide using synthetic division $$ ( x^{4}-9x^{3}-4x^{2}+3 ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}2&1&-9&-4&0&3\\& & 2& -14& -36& \color{black}{-72} \\ \hline &\color{blue}{1}&\color{blue}{-7}&\color{blue}{-18}&\color{blue}{-36}&\color{orangered}{-69} \end{array} $$The solution is:
$$ \frac{ x^{4}-9x^{3}-4x^{2}+3 }{ x-2 } = \color{blue}{x^{3}-7x^{2}-18x-36} \color{red}{~-~} \frac{ \color{red}{ 69 } }{ x-2 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -2 = 0 $ ( $ x = \color{blue}{ 2 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&-9&-4&0&3\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}2&\color{orangered}{ 1 }&-9&-4&0&3\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 1 } = \color{blue}{ 2 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&-9&-4&0&3\\& & \color{blue}{2} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -9 } + \color{orangered}{ 2 } = \color{orangered}{ -7 } $
$$ \begin{array}{c|rrrrr}2&1&\color{orangered}{ -9 }&-4&0&3\\& & \color{orangered}{2} & & & \\ \hline &1&\color{orangered}{-7}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -7 \right) } = \color{blue}{ -14 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&-9&-4&0&3\\& & 2& \color{blue}{-14} & & \\ \hline &1&\color{blue}{-7}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ \left( -14 \right) } = \color{orangered}{ -18 } $
$$ \begin{array}{c|rrrrr}2&1&-9&\color{orangered}{ -4 }&0&3\\& & 2& \color{orangered}{-14} & & \\ \hline &1&-7&\color{orangered}{-18}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -18 \right) } = \color{blue}{ -36 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&-9&-4&0&3\\& & 2& -14& \color{blue}{-36} & \\ \hline &1&-7&\color{blue}{-18}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -36 \right) } = \color{orangered}{ -36 } $
$$ \begin{array}{c|rrrrr}2&1&-9&-4&\color{orangered}{ 0 }&3\\& & 2& -14& \color{orangered}{-36} & \\ \hline &1&-7&-18&\color{orangered}{-36}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -36 \right) } = \color{blue}{ -72 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&-9&-4&0&3\\& & 2& -14& -36& \color{blue}{-72} \\ \hline &1&-7&-18&\color{blue}{-36}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ \left( -72 \right) } = \color{orangered}{ -69 } $
$$ \begin{array}{c|rrrrr}2&1&-9&-4&0&\color{orangered}{ 3 }\\& & 2& -14& -36& \color{orangered}{-72} \\ \hline &\color{blue}{1}&\color{blue}{-7}&\color{blue}{-18}&\color{blue}{-36}&\color{orangered}{-69} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}-7x^{2}-18x-36 } $ with a remainder of $ \color{red}{ -69 } $.