Divide using synthetic division $$ ( 2x^{4}-5x^{3}+x+3 ) \div ( 2x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}3&2&-5&0&1&3\\& & 6& 3& 9& \color{black}{30} \\ \hline &\color{blue}{2}&\color{blue}{1}&\color{blue}{3}&\color{blue}{10}&\color{orangered}{33} \end{array} $$The solution is:
$$ \frac{ 2x^{4}-5x^{3}+x+3 }{ x-3 } = \color{blue}{2x^{3}+x^{2}+3x+10} ~+~ \frac{ \color{red}{ 33 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&-5&0&1&3\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}3&\color{orangered}{ 2 }&-5&0&1&3\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&-5&0&1&3\\& & \color{blue}{6} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ 6 } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrrr}3&2&\color{orangered}{ -5 }&0&1&3\\& & \color{orangered}{6} & & & \\ \hline &2&\color{orangered}{1}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 1 } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&-5&0&1&3\\& & 6& \color{blue}{3} & & \\ \hline &2&\color{blue}{1}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 3 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrrr}3&2&-5&\color{orangered}{ 0 }&1&3\\& & 6& \color{orangered}{3} & & \\ \hline &2&1&\color{orangered}{3}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 3 } = \color{blue}{ 9 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&-5&0&1&3\\& & 6& 3& \color{blue}{9} & \\ \hline &2&1&\color{blue}{3}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 9 } = \color{orangered}{ 10 } $
$$ \begin{array}{c|rrrrr}3&2&-5&0&\color{orangered}{ 1 }&3\\& & 6& 3& \color{orangered}{9} & \\ \hline &2&1&3&\color{orangered}{10}& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&-5&0&1&3\\& & 6& 3& 9& \color{blue}{30} \\ \hline &2&1&3&\color{blue}{10}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 30 } = \color{orangered}{ 33 } $
$$ \begin{array}{c|rrrrr}3&2&-5&0&1&\color{orangered}{ 3 }\\& & 6& 3& 9& \color{orangered}{30} \\ \hline &\color{blue}{2}&\color{blue}{1}&\color{blue}{3}&\color{blue}{10}&\color{orangered}{33} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{3}+x^{2}+3x+10 } $ with a remainder of $ \color{red}{ 33 } $.