Divide using synthetic division $$ ( 4x^{5}+11x^{3}-x^{2}+9 ) \div ( 2x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}3&4&0&11&-1&0&9\\& & 12& 36& 141& 420& \color{black}{1260} \\ \hline &\color{blue}{4}&\color{blue}{12}&\color{blue}{47}&\color{blue}{140}&\color{blue}{420}&\color{orangered}{1269} \end{array} $$The solution is:
$$ \frac{ 4x^{5}+11x^{3}-x^{2}+9 }{ x-3 } = \color{blue}{4x^{4}+12x^{3}+47x^{2}+140x+420} ~+~ \frac{ \color{red}{ 1269 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&4&0&11&-1&0&9\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}3&\color{orangered}{ 4 }&0&11&-1&0&9\\& & & & & & \\ \hline &\color{orangered}{4}&&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 4 } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&4&0&11&-1&0&9\\& & \color{blue}{12} & & & & \\ \hline &\color{blue}{4}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 12 } = \color{orangered}{ 12 } $
$$ \begin{array}{c|rrrrrr}3&4&\color{orangered}{ 0 }&11&-1&0&9\\& & \color{orangered}{12} & & & & \\ \hline &4&\color{orangered}{12}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 12 } = \color{blue}{ 36 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&4&0&11&-1&0&9\\& & 12& \color{blue}{36} & & & \\ \hline &4&\color{blue}{12}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 11 } + \color{orangered}{ 36 } = \color{orangered}{ 47 } $
$$ \begin{array}{c|rrrrrr}3&4&0&\color{orangered}{ 11 }&-1&0&9\\& & 12& \color{orangered}{36} & & & \\ \hline &4&12&\color{orangered}{47}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 47 } = \color{blue}{ 141 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&4&0&11&-1&0&9\\& & 12& 36& \color{blue}{141} & & \\ \hline &4&12&\color{blue}{47}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 141 } = \color{orangered}{ 140 } $
$$ \begin{array}{c|rrrrrr}3&4&0&11&\color{orangered}{ -1 }&0&9\\& & 12& 36& \color{orangered}{141} & & \\ \hline &4&12&47&\color{orangered}{140}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 140 } = \color{blue}{ 420 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&4&0&11&-1&0&9\\& & 12& 36& 141& \color{blue}{420} & \\ \hline &4&12&47&\color{blue}{140}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 420 } = \color{orangered}{ 420 } $
$$ \begin{array}{c|rrrrrr}3&4&0&11&-1&\color{orangered}{ 0 }&9\\& & 12& 36& 141& \color{orangered}{420} & \\ \hline &4&12&47&140&\color{orangered}{420}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 420 } = \color{blue}{ 1260 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&4&0&11&-1&0&9\\& & 12& 36& 141& 420& \color{blue}{1260} \\ \hline &4&12&47&140&\color{blue}{420}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ 9 } + \color{orangered}{ 1260 } = \color{orangered}{ 1269 } $
$$ \begin{array}{c|rrrrrr}3&4&0&11&-1&0&\color{orangered}{ 9 }\\& & 12& 36& 141& 420& \color{orangered}{1260} \\ \hline &\color{blue}{4}&\color{blue}{12}&\color{blue}{47}&\color{blue}{140}&\color{blue}{420}&\color{orangered}{1269} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{4}+12x^{3}+47x^{2}+140x+420 } $ with a remainder of $ \color{red}{ 1269 } $.