Divide using synthetic division $$ ( 2x^{4}-12x^{3}+3x-9 ) \div ( x-6 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}6&2&-12&0&3&-9\\& & 12& 0& 0& \color{black}{18} \\ \hline &\color{blue}{2}&\color{blue}{0}&\color{blue}{0}&\color{blue}{3}&\color{orangered}{9} \end{array} $$The solution is:
$$ \frac{ 2x^{4}-12x^{3}+3x-9 }{ x-6 } = \color{blue}{2x^{3}+3} ~+~ \frac{ \color{red}{ 9 } }{ x-6 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -6 = 0 $ ( $ x = \color{blue}{ 6 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{6}&2&-12&0&3&-9\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}6&\color{orangered}{ 2 }&-12&0&3&-9\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 6 } \cdot \color{blue}{ 2 } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{6}&2&-12&0&3&-9\\& & \color{blue}{12} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -12 } + \color{orangered}{ 12 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}6&2&\color{orangered}{ -12 }&0&3&-9\\& & \color{orangered}{12} & & & \\ \hline &2&\color{orangered}{0}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 6 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{6}&2&-12&0&3&-9\\& & 12& \color{blue}{0} & & \\ \hline &2&\color{blue}{0}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 0 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}6&2&-12&\color{orangered}{ 0 }&3&-9\\& & 12& \color{orangered}{0} & & \\ \hline &2&0&\color{orangered}{0}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 6 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{6}&2&-12&0&3&-9\\& & 12& 0& \color{blue}{0} & \\ \hline &2&0&\color{blue}{0}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 0 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrrr}6&2&-12&0&\color{orangered}{ 3 }&-9\\& & 12& 0& \color{orangered}{0} & \\ \hline &2&0&0&\color{orangered}{3}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 6 } \cdot \color{blue}{ 3 } = \color{blue}{ 18 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{6}&2&-12&0&3&-9\\& & 12& 0& 0& \color{blue}{18} \\ \hline &2&0&0&\color{blue}{3}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -9 } + \color{orangered}{ 18 } = \color{orangered}{ 9 } $
$$ \begin{array}{c|rrrrr}6&2&-12&0&3&\color{orangered}{ -9 }\\& & 12& 0& 0& \color{orangered}{18} \\ \hline &\color{blue}{2}&\color{blue}{0}&\color{blue}{0}&\color{blue}{3}&\color{orangered}{9} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{3}+3 } $ with a remainder of $ \color{red}{ 9 } $.