Divide using synthetic division $$ ( 2x^{3}+2x^{2}-31x+21 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}3&2&2&-31&21\\& & 6& 24& \color{black}{-21} \\ \hline &\color{blue}{2}&\color{blue}{8}&\color{blue}{-7}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 2x^{3}+2x^{2}-31x+21 }{ x-3 } = \color{blue}{2x^{2}+8x-7} $$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.
$$ \begin{array}{c|rrrr}\color{blue}{3}&2&2&-31&21\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}3&\color{orangered}{ 2 }&2&-31&21\\& & & & \\ \hline &\color{orangered}{2}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&2&2&-31&21\\& & \color{blue}{6} & & \\ \hline &\color{blue}{2}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 6 } = \color{orangered}{ 8 } $
$$ \begin{array}{c|rrrr}3&2&\color{orangered}{ 2 }&-31&21\\& & \color{orangered}{6} & & \\ \hline &2&\color{orangered}{8}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 8 } = \color{blue}{ 24 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&2&2&-31&21\\& & 6& \color{blue}{24} & \\ \hline &2&\color{blue}{8}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -31 } + \color{orangered}{ 24 } = \color{orangered}{ -7 } $
$$ \begin{array}{c|rrrr}3&2&2&\color{orangered}{ -31 }&21\\& & 6& \color{orangered}{24} & \\ \hline &2&8&\color{orangered}{-7}& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&2&2&-31&21\\& & 6& 24& \color{blue}{-21} \\ \hline &2&8&\color{blue}{-7}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 21 } + \color{orangered}{ \left( -21 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}3&2&2&-31&\color{orangered}{ 21 }\\& & 6& 24& \color{orangered}{-21} \\ \hline &\color{blue}{2}&\color{blue}{8}&\color{blue}{-7}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{2}+8x-7 } $ with a remainder of $ \color{red}{ 0 } $.