Divide using synthetic division $$ ( 3x^{4}-2x+5 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}3&3&0&0&-2&5\\& & 9& 27& 81& \color{black}{237} \\ \hline &\color{blue}{3}&\color{blue}{9}&\color{blue}{27}&\color{blue}{79}&\color{orangered}{242} \end{array} $$The solution is:
$$ \frac{ 3x^{4}-2x+5 }{ x-3 } = \color{blue}{3x^{3}+9x^{2}+27x+79} ~+~ \frac{ \color{red}{ 242 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&3&0&0&-2&5\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}3&\color{orangered}{ 3 }&0&0&-2&5\\& & & & & \\ \hline &\color{orangered}{3}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 3 } = \color{blue}{ 9 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&3&0&0&-2&5\\& & \color{blue}{9} & & & \\ \hline &\color{blue}{3}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 9 } = \color{orangered}{ 9 } $
$$ \begin{array}{c|rrrrr}3&3&\color{orangered}{ 0 }&0&-2&5\\& & \color{orangered}{9} & & & \\ \hline &3&\color{orangered}{9}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 9 } = \color{blue}{ 27 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&3&0&0&-2&5\\& & 9& \color{blue}{27} & & \\ \hline &3&\color{blue}{9}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 27 } = \color{orangered}{ 27 } $
$$ \begin{array}{c|rrrrr}3&3&0&\color{orangered}{ 0 }&-2&5\\& & 9& \color{orangered}{27} & & \\ \hline &3&9&\color{orangered}{27}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 27 } = \color{blue}{ 81 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&3&0&0&-2&5\\& & 9& 27& \color{blue}{81} & \\ \hline &3&9&\color{blue}{27}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ 81 } = \color{orangered}{ 79 } $
$$ \begin{array}{c|rrrrr}3&3&0&0&\color{orangered}{ -2 }&5\\& & 9& 27& \color{orangered}{81} & \\ \hline &3&9&27&\color{orangered}{79}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 79 } = \color{blue}{ 237 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&3&0&0&-2&5\\& & 9& 27& 81& \color{blue}{237} \\ \hline &3&9&27&\color{blue}{79}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ 237 } = \color{orangered}{ 242 } $
$$ \begin{array}{c|rrrrr}3&3&0&0&-2&\color{orangered}{ 5 }\\& & 9& 27& 81& \color{orangered}{237} \\ \hline &\color{blue}{3}&\color{blue}{9}&\color{blue}{27}&\color{blue}{79}&\color{orangered}{242} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{3}+9x^{2}+27x+79 } $ with a remainder of $ \color{red}{ 242 } $.