Divide using synthetic division $$ ( 2x^{4}-14x^{3}+28x^{2}-5x-21 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}3&2&-14&28&-5&-21\\& & 6& -24& 12& \color{black}{21} \\ \hline &\color{blue}{2}&\color{blue}{-8}&\color{blue}{4}&\color{blue}{7}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 2x^{4}-14x^{3}+28x^{2}-5x-21 }{ x-3 } = \color{blue}{2x^{3}-8x^{2}+4x+7} $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&-14&28&-5&-21\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}3&\color{orangered}{ 2 }&-14&28&-5&-21\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&-14&28&-5&-21\\& & \color{blue}{6} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -14 } + \color{orangered}{ 6 } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrrr}3&2&\color{orangered}{ -14 }&28&-5&-21\\& & \color{orangered}{6} & & & \\ \hline &2&\color{orangered}{-8}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -8 \right) } = \color{blue}{ -24 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&-14&28&-5&-21\\& & 6& \color{blue}{-24} & & \\ \hline &2&\color{blue}{-8}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 28 } + \color{orangered}{ \left( -24 \right) } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrrr}3&2&-14&\color{orangered}{ 28 }&-5&-21\\& & 6& \color{orangered}{-24} & & \\ \hline &2&-8&\color{orangered}{4}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 4 } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&-14&28&-5&-21\\& & 6& -24& \color{blue}{12} & \\ \hline &2&-8&\color{blue}{4}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ 12 } = \color{orangered}{ 7 } $
$$ \begin{array}{c|rrrrr}3&2&-14&28&\color{orangered}{ -5 }&-21\\& & 6& -24& \color{orangered}{12} & \\ \hline &2&-8&4&\color{orangered}{7}& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&-14&28&-5&-21\\& & 6& -24& 12& \color{blue}{21} \\ \hline &2&-8&4&\color{blue}{7}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -21 } + \color{orangered}{ 21 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}3&2&-14&28&-5&\color{orangered}{ -21 }\\& & 6& -24& 12& \color{orangered}{21} \\ \hline &\color{blue}{2}&\color{blue}{-8}&\color{blue}{4}&\color{blue}{7}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{3}-8x^{2}+4x+7 } $ with a remainder of $ \color{red}{ 0 } $.