Divide using synthetic division $$ ( 2x^{4}-9x^{3}+21x^{2}-26x+12 ) \div ( 2x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}3&2&-9&21&-26&12\\& & 6& -9& 36& \color{black}{30} \\ \hline &\color{blue}{2}&\color{blue}{-3}&\color{blue}{12}&\color{blue}{10}&\color{orangered}{42} \end{array} $$The solution is:
$$ \frac{ 2x^{4}-9x^{3}+21x^{2}-26x+12 }{ x-3 } = \color{blue}{2x^{3}-3x^{2}+12x+10} ~+~ \frac{ \color{red}{ 42 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&-9&21&-26&12\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}3&\color{orangered}{ 2 }&-9&21&-26&12\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&-9&21&-26&12\\& & \color{blue}{6} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -9 } + \color{orangered}{ 6 } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrrr}3&2&\color{orangered}{ -9 }&21&-26&12\\& & \color{orangered}{6} & & & \\ \hline &2&\color{orangered}{-3}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ -9 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&-9&21&-26&12\\& & 6& \color{blue}{-9} & & \\ \hline &2&\color{blue}{-3}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 21 } + \color{orangered}{ \left( -9 \right) } = \color{orangered}{ 12 } $
$$ \begin{array}{c|rrrrr}3&2&-9&\color{orangered}{ 21 }&-26&12\\& & 6& \color{orangered}{-9} & & \\ \hline &2&-3&\color{orangered}{12}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 12 } = \color{blue}{ 36 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&-9&21&-26&12\\& & 6& -9& \color{blue}{36} & \\ \hline &2&-3&\color{blue}{12}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -26 } + \color{orangered}{ 36 } = \color{orangered}{ 10 } $
$$ \begin{array}{c|rrrrr}3&2&-9&21&\color{orangered}{ -26 }&12\\& & 6& -9& \color{orangered}{36} & \\ \hline &2&-3&12&\color{orangered}{10}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 10 } = \color{blue}{ 30 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&-9&21&-26&12\\& & 6& -9& 36& \color{blue}{30} \\ \hline &2&-3&12&\color{blue}{10}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ 30 } = \color{orangered}{ 42 } $
$$ \begin{array}{c|rrrrr}3&2&-9&21&-26&\color{orangered}{ 12 }\\& & 6& -9& 36& \color{orangered}{30} \\ \hline &\color{blue}{2}&\color{blue}{-3}&\color{blue}{12}&\color{blue}{10}&\color{orangered}{42} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{3}-3x^{2}+12x+10 } $ with a remainder of $ \color{red}{ 42 } $.