Divide using synthetic division $$ ( 2x^{4}-8x^{3}-3x^{2}+28x-3 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}3&2&-8&-3&28&-3\\& & 6& -6& -27& \color{black}{3} \\ \hline &\color{blue}{2}&\color{blue}{-2}&\color{blue}{-9}&\color{blue}{1}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 2x^{4}-8x^{3}-3x^{2}+28x-3 }{ x-3 } = \color{blue}{2x^{3}-2x^{2}-9x+1} $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&-8&-3&28&-3\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}3&\color{orangered}{ 2 }&-8&-3&28&-3\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&-8&-3&28&-3\\& & \color{blue}{6} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -8 } + \color{orangered}{ 6 } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrrr}3&2&\color{orangered}{ -8 }&-3&28&-3\\& & \color{orangered}{6} & & & \\ \hline &2&\color{orangered}{-2}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ -6 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&-8&-3&28&-3\\& & 6& \color{blue}{-6} & & \\ \hline &2&\color{blue}{-2}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ -9 } $
$$ \begin{array}{c|rrrrr}3&2&-8&\color{orangered}{ -3 }&28&-3\\& & 6& \color{orangered}{-6} & & \\ \hline &2&-2&\color{orangered}{-9}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -9 \right) } = \color{blue}{ -27 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&-8&-3&28&-3\\& & 6& -6& \color{blue}{-27} & \\ \hline &2&-2&\color{blue}{-9}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 28 } + \color{orangered}{ \left( -27 \right) } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrrr}3&2&-8&-3&\color{orangered}{ 28 }&-3\\& & 6& -6& \color{orangered}{-27} & \\ \hline &2&-2&-9&\color{orangered}{1}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 1 } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&-8&-3&28&-3\\& & 6& -6& -27& \color{blue}{3} \\ \hline &2&-2&-9&\color{blue}{1}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ 3 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}3&2&-8&-3&28&\color{orangered}{ -3 }\\& & 6& -6& -27& \color{orangered}{3} \\ \hline &\color{blue}{2}&\color{blue}{-2}&\color{blue}{-9}&\color{blue}{1}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{3}-2x^{2}-9x+1 } $ with a remainder of $ \color{red}{ 0 } $.