Divide using synthetic division $$ ( 2x^{4}-15x^{3}+20x^{2}+30x-30 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}3&2&-15&20&30&-30\\& & 6& -27& -21& \color{black}{27} \\ \hline &\color{blue}{2}&\color{blue}{-9}&\color{blue}{-7}&\color{blue}{9}&\color{orangered}{-3} \end{array} $$The solution is:
$$ \frac{ 2x^{4}-15x^{3}+20x^{2}+30x-30 }{ x-3 } = \color{blue}{2x^{3}-9x^{2}-7x+9} \color{red}{~-~} \frac{ \color{red}{ 3 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&-15&20&30&-30\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}3&\color{orangered}{ 2 }&-15&20&30&-30\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&-15&20&30&-30\\& & \color{blue}{6} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -15 } + \color{orangered}{ 6 } = \color{orangered}{ -9 } $
$$ \begin{array}{c|rrrrr}3&2&\color{orangered}{ -15 }&20&30&-30\\& & \color{orangered}{6} & & & \\ \hline &2&\color{orangered}{-9}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -9 \right) } = \color{blue}{ -27 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&-15&20&30&-30\\& & 6& \color{blue}{-27} & & \\ \hline &2&\color{blue}{-9}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 20 } + \color{orangered}{ \left( -27 \right) } = \color{orangered}{ -7 } $
$$ \begin{array}{c|rrrrr}3&2&-15&\color{orangered}{ 20 }&30&-30\\& & 6& \color{orangered}{-27} & & \\ \hline &2&-9&\color{orangered}{-7}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -7 \right) } = \color{blue}{ -21 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&-15&20&30&-30\\& & 6& -27& \color{blue}{-21} & \\ \hline &2&-9&\color{blue}{-7}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 30 } + \color{orangered}{ \left( -21 \right) } = \color{orangered}{ 9 } $
$$ \begin{array}{c|rrrrr}3&2&-15&20&\color{orangered}{ 30 }&-30\\& & 6& -27& \color{orangered}{-21} & \\ \hline &2&-9&-7&\color{orangered}{9}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 9 } = \color{blue}{ 27 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&-15&20&30&-30\\& & 6& -27& -21& \color{blue}{27} \\ \hline &2&-9&-7&\color{blue}{9}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -30 } + \color{orangered}{ 27 } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrrr}3&2&-15&20&30&\color{orangered}{ -30 }\\& & 6& -27& -21& \color{orangered}{27} \\ \hline &\color{blue}{2}&\color{blue}{-9}&\color{blue}{-7}&\color{blue}{9}&\color{orangered}{-3} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{3}-9x^{2}-7x+9 } $ with a remainder of $ \color{red}{ -3 } $.