Divide using synthetic division $$ ( 2x^{4}-16x^{2}+x-6 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}3&2&0&-16&1&-6\\& & 6& 18& 6& \color{black}{21} \\ \hline &\color{blue}{2}&\color{blue}{6}&\color{blue}{2}&\color{blue}{7}&\color{orangered}{15} \end{array} $$The solution is:
$$ \frac{ 2x^{4}-16x^{2}+x-6 }{ x-3 } = \color{blue}{2x^{3}+6x^{2}+2x+7} ~+~ \frac{ \color{red}{ 15 } }{ x-3 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -3 = 0 $ ( $ x = \color{blue}{ 3 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&0&-16&1&-6\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}3&\color{orangered}{ 2 }&0&-16&1&-6\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 2 } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&0&-16&1&-6\\& & \color{blue}{6} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 6 } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrrr}3&2&\color{orangered}{ 0 }&-16&1&-6\\& & \color{orangered}{6} & & & \\ \hline &2&\color{orangered}{6}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 6 } = \color{blue}{ 18 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&0&-16&1&-6\\& & 6& \color{blue}{18} & & \\ \hline &2&\color{blue}{6}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -16 } + \color{orangered}{ 18 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrr}3&2&0&\color{orangered}{ -16 }&1&-6\\& & 6& \color{orangered}{18} & & \\ \hline &2&6&\color{orangered}{2}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 2 } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&0&-16&1&-6\\& & 6& 18& \color{blue}{6} & \\ \hline &2&6&\color{blue}{2}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 6 } = \color{orangered}{ 7 } $
$$ \begin{array}{c|rrrrr}3&2&0&-16&\color{orangered}{ 1 }&-6\\& & 6& 18& \color{orangered}{6} & \\ \hline &2&6&2&\color{orangered}{7}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 7 } = \color{blue}{ 21 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&0&-16&1&-6\\& & 6& 18& 6& \color{blue}{21} \\ \hline &2&6&2&\color{blue}{7}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ 21 } = \color{orangered}{ 15 } $
$$ \begin{array}{c|rrrrr}3&2&0&-16&1&\color{orangered}{ -6 }\\& & 6& 18& 6& \color{orangered}{21} \\ \hline &\color{blue}{2}&\color{blue}{6}&\color{blue}{2}&\color{blue}{7}&\color{orangered}{15} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{3}+6x^{2}+2x+7 } $ with a remainder of $ \color{red}{ 15 } $.